Looking at this problem, I need to find the largest possible median after repeatedly removing subarrays of size k until the array has exactly k elements remaining.

Let me analyze the structure:
- We start with n elements, and keep removing k elements until we have exactly k elements left (since we stop when |a| > k is false, meaning |a| <= k... wait, we stop when |a| is NOT greater than k, so we stop when |a| <= k. But each operation removes k elements. We need n - m*k = k for some m, so n = (m+1)*k, meaning n must be divisible by k? Let me re-read.

Actually "repeat while |a| > k". So we stop when |a| <= k. But since each removal decreases size by k, if we start with n and do m operations, size becomes n - m*k. We stop when n - m*k <= k, i.e., m >= (n-k)/k. So final size is n mod k if n mod k != 0, else k. Wait: final size is ((n-k) mod k) + ... let me think. Final size r where r = n mod k if n mod k != 0, else k. Actually r = n - floor((n-k)/k)*k... Let me just compute: we do as many operations as possible. Each removes k. We stop when size <= k. So final size = n - floor((n-1)/k - ... hmm.

Final size: we want largest m such that n - mk > k, i.e., m < (n-k)/k. So m = floor((n-k-1)/k) = floor((n-1)/k) - 1. Final size = n - mk. Let r = n mod k; final size = r if r != 0 else k. Wait: if n = 4, k = 3: r = 1, but answer array has size 1? No wait example 1: n=4, k=3, final size should be 1 (we do 1 operation). Yes, [3] is size 1.

The median of the final array of size r is at position floor((r+1)/2).

Now binary search on the answer: can we achieve median >= x? 

For a fixed x, mark elements as 1 if >= x, else -1. We want to maximize the count of 1s (elements >= x) in the remaining array. The median >= x iff at least ceil((r+1)/2) = floor((r+1)/2) elements are >= x... actually median >= x means the element at position floor((r+1)/2) is >= x, which means at least (r - floor((r+1)/2) + 1) elements are >= x. Let pos = floor((r+1)/2), need at least (r - pos + 1) elements >= x.

Binary search: for a given x, can we select k remaining elements (or r elements) such that the number >= x is maximized?

We need to choose which elements to keep. The elements we keep must be "selectable" - we remove contiguous blocks of k from the original array. The remaining elements form a subsequence that can be obtained by removing some set of non-overlapping subarrays of size k.

Key insight: which subsequences of size r = (n mod k == 0 ? k : n mod k) are achievable? The remaining elements must be obtainable by removing blocks. 

The remaining indices must satisfy: if we sort them as i_1 < i_2 < ... < i_r, the gaps must be fillable with complete blocks of k. The number of removed elements between positions must be multiples of k (including before first and after last).

So constraints: i_1 - 1 ≡ 0 (mod k), meaning i_1 ≡ 1 (mod k). Between consecutive kept elements: i_{j+1} - i_j - 1 ≡ 0 (mod k). After last: n - i_r ≡ 0 (mod k).

So i_1 ≡ 1 (mod k), each gap i_{j+1} - i_j ≡ 1 (mod k), and n - i_r ≡ 0 (mod k), i.e., i_r ≡ n (mod k).

This means all kept indices i_j satisfy i_j ≡ j (mod k) (1-indexed).

So kept element j (1-indexed) must come from index ≡ j (mod k).

To maximize elements >= x: for each position j in 1..r, pick index ≡ j (mod k) with value >= x if possible, and we need them ordered.

This is a selection problem with DP or greedy.

```go
package main

import (
	"bufio"
	"fmt"
	"os"
)

func main() {
	reader := bufio.NewReader(os.Stdin)
	writer := bufio.NewWriter(os.Stdout)
	defer writer.Flush()
	var t int
	fmt.Fscan(reader, &t)
	for ; t > 0; t-- {
		var n, k int
		fmt.Fscan(reader, &n, &k)
		a := make([]int, n)
		for i := range a { fmt.Fscan(reader, &a[i]) }
		r := n % k; if r == 0 { r = k }
		// kept element j (1..r) must have index i where (i-1)%k == (j-1)%k, i.e., i%k == j%k (with 1-indexed)
		// Actually i_j ≡ j (mod k), 1-indexed
		// Binary search on answer
		canAchieve := func(x int) bool {
			// For each position j in 1..r, find best index with i≡j(mod k), ordered
			// Greedy: for each j, among indices with i≡j(mod k) and i > prev, pick one maximizing score
			// Score: 1 if a[i]>=x else 0. We want total >= need
			need := r - (r+1)/2 + 1 // number of elements >= x needed
			// DP: dp[j] = max elements >= x in first j kept positions
			// For position j, valid indices are j, j+k, j+2k, ... (1-indexed)
			// We need prev_index < cur_index, and cur_index ≡ j (mod k)
			// Greedy: track last chosen index, for position j choose smallest valid index > prev that maximizes
			// Actually we want to maximize count, so: for each j, if we can pick index >= x, do so with earliest such index > prev; else pick earliest index > prev
			prev := 0
			count := 0
			for j := 1; j <= r; j++ {
				// indices: j, j+k, j+2k, ... all <= n, 1-indexed
				// find smallest index > prev with i≡j(mod k)
				// i = j + m*k > prev => m > (prev-j)/k => m = max(0, floor((prev-j)/k)+1)
				var start int
				if prev < j { start = j } else {
					m := (prev - j) / k + 1
					start = j + m*k
				}
				if start > n { return false }
				// Now find if any index in [start, start+k-1, start+2k,...] <= n has a[i] >= x
				// To maximize, we want to pick >= x if possible, earliest
				// But we also need to leave room for remaining r-j positions
				// Remaining after position j: r-j more, each needs at least one slot
				// Last index for position j: n - (r-j)*k (since remaining r-j need indices j+1+...(r-j)*k apart minimum)
				// Actually max index for position j: n - (r-j)*k... but step is k
				// max index for pos j with i≡j(mod k): let's compute
				// last possible: must leave r-j indices after, each ≡ j+1,...,r (mod k) and >cur
				// min space needed: (r-j)*1 but with mod constraint each subsequent +k minimum
				// so max_index for pos j = n - (r-j)*k
				maxIdx := n - (r-j)*k
				if maxIdx < start { return false }
				// Find earliest index in [start..maxIdx] step k with a[i]>=x
				foundGood := -1
				for idx := start; idx <= maxIdx; idx += k {
					if a[idx-1] >= x { foundGood = idx; break }
				}
				if foundGood != -1 { count++; prev = foundGood } else { prev = start }
			}
			return count >= need
		}
		// collect unique values, binary search
		vals := make([]int, n)
		copy(vals, a)
		// sort and unique
		// binary search: largest x where canAchieve(x)
		lo, hi := 0, n-1
		// sort a for binary search on values
		sorted := make([]int, n)
		copy(sorted, a)
		// sort sorted
		for i := 1; i < n; i++ {
			v := sorted[i]; j := i-1
			for j >= 0 && sorted[j] > v { sorted[j+1] = sorted[j]; j-- }
			sorted[j+1] = v
		}
		_ = vals
		ans := sorted[0]
		for lo <= hi {
			mid := (lo+hi)/2
			if canAchieve(sorted[mid]) { ans = sorted[mid]; lo = mid+1 } else { hi = mid-1 }
		}
		fmt.Fprintln(writer, ans)
	}
}
```

Wait, the inner loop in canAchieve can be O(n/k * r) = O(n) per call, and binary search is O(log n), so O(n log n) total. But with t=10^4 and sum n=5*10^5, this should be fine.

But insertion sort is O(n^2)! Let me use a proper sort.

```go
package main

import (
	"bufio"
	"fmt"
	"os"
	"sort"
)

func main() {
	reader := bufio.NewReader(os.Stdin)
	writer := bufio.NewWriter(os.Stdout)
	defer writer.Flush()
	var t int
	fmt.Fscan(reader, &t)
	for ; t > 0; t-- {
		var n, k int
		fmt.Fscan(reader, &n, &k)
		a := make([]int, n)
		for i := range a {
			fmt.Fscan(reader, &a[i])
		}
		r := n % k
		if r == 0 {
			r = k
		}
		need := r - (r+1)/2 + 1

		canAchieve := func(x int) bool {
			prev := 0
			count := 0
			for j := 1; j <= r; j++ {
				var start int
				if prev < j {
					start = j
				} else {
					m := (prev-j)/k + 1
					start = j + m*k
				}
				if start > n {
					return false
				}
				maxIdx := n - (r-j)*k
				if maxIdx < start {
					return false
				}
				foundGood := -1
				for idx := start; idx <= maxIdx; idx += k {
					if a[idx-1] >= x {
						foundGood = idx
						break
					}
				}
				if foundGood != -1 {
					count++
					prev = foundGood
				} else {
					prev = start
				}
			}
			return count >= need
		}

		sorted := make([]int, n)
		copy(sorted, a)
		sort.Ints(sorted)

		lo, hi := 0, n-1
		ans := sorted[0]
		for lo <= hi {
			mid := (lo + hi) / 2
			if canAchieve(sorted[mid]) {
				ans = sorted[mid]
				lo = mid + 1
			} else {
				hi = mid - 1
			}
		}
		fmt.Fprintln(writer, ans)
	}
}
```

Let me trace through example 1: n=4, k=3, a=[3,9,9,2], r=1, need=1.
canAchieve(x): j=1, prev=0, start=1, maxIdx=4-0=4. Find idx in {1,4} with a[idx-1]>=x.
For x=9: a[0]=3<9, a[3]=2<9. foundGood=-1, prev=1. count=0 < 1. return false.
For x=3: a[0]=3>=3. foundGood=1, count=1>=1. return true. ans=3. Correct!

Example 2: n=5,k=3,a=[3,2,5,6,4], r=2, need=1.
canAchieve(6): j=1,prev=0,start=1,maxIdx=5-3=2. idx=1: a[0]=3<6. idx... wait step is k=3, so idx=1,4 but maxIdx=2. Only idx=1. foundGood=-1, prev=1. j=2,prev=1,start=2,maxIdx=5. idx=2: a[1]=2<6. idx=5: a[4]=4<6. foundGood=-1, prev=2. count=0<1. false.

Hmm but answer is 4. Let me try canAchieve(4):
j=1: start=1, maxIdx=2. idx=1: a[0]=3<4. foundGood=-1, prev=1.
j=2: start=2, maxIdx=5. idx=2: a[1]=2<4. idx=5: a[4]=4>=4. foundGood=5, count=1>=1. return true.

So ans=4. Correct!

Let me check example 3: n=7,k=1,a=[5,9,2,6,5,4,6], r=7%1=0 so r=1, need=1.
canAchieve(9): j=1, start=1, maxIdx=7. idx=1,2,...,7 step 1. a[1]=9>=9. foundGood=2? wait idx starts at 1, a[0]=5<9, idx=2, a[1]=9>=9. foundGood=2, count=1>=1. true. ans=9. Correct!

The solution looks correct. Let me also verify the complexity concern: for each binary search step, canAchieve runs in O(n) since total iterations across all j is at most n (each idx increment moves by k, and we have r groups, total range n). So O(n log n) per test case total.