For problem statement at 1000-1999/1000-1099/1090-1099/1098/problemC.txt this is a correct solution, but verifier at 1000-1999/1000-1099/1090-1099/1098/verifierC.go ends with Wrong answer on test 1
Input:
9 35
Expected:
Yes
1 1 2 4 5 5 6 8
Got:
Yes
1 1 2 9 7 8 9 4 

exit status 1 can you fix the verifier? ```go
package main

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

func main() {
	reader := bufio.NewReader(os.Stdin)
	writer := bufio.NewWriter(os.Stdout)
	defer writer.Flush()

	var n int
	var s int64
	fmt.Fscan(reader, &n, &s)

	// Check feasibility range
	// Min sum is achieved by a star graph (branching coefficient n-1): 1 (root) + 2*(n-1) = 2n-1
	// Max sum is achieved by a path (branching coefficient 1): n*(n+1)/2
	if s < int64(2*n-1) || s > int64(n)*int64(n+1)/2 {
		fmt.Fprintln(writer, "No")
		return
	}

	// Function to calculate minimum sum of subtree sizes (depths) for a given branching coefficient k
	calcMinSum := func(k int) int64 {
		if k == 1 {
			return int64(n) * int64(n+1) / 2
		}
		var sum int64 = 0
		rem := n
		depth := 1
		count := 1 // nodes at current depth
		for rem > 0 {
			take := rem
			if take > count {
				take = count
			}
			sum += int64(take) * int64(depth)
			rem -= take

			// Calculate capacity for next level
			// count * k can overflow int if not careful, though n <= 10^5 makes it safe in int64
			nextCount := int64(count) * int64(k)
			if nextCount > int64(n) {
				count = n + 1 // Cap at > n
			} else {
				count = int(nextCount)
			}
			depth++
		}
		return sum
	}

	// Binary search for the minimum branching coefficient k
	low := 1
	high := n - 1
	ansK := n - 1

	for low <= high {
		mid := (low + high) / 2
		if calcMinSum(mid) <= s {
			ansK = mid
			high = mid - 1
		} else {
			low = mid + 1
		}
	}

	fmt.Fprintln(writer, "Yes")

	// Construct the parent array for the complete k-ary tree (BFS order)
	p := make([]int, n+1)
	d := make([]int, n+1)
	degree := make([]int, n+1)

	d[1] = 1
	var currSum int64 = 1
	// Standard BFS parent assignment: node i (1-based)
	// For k-ary tree rooted at 1:
	// Children of 1 are 2..k+1
	// Children of 2 are k+2..2k+1, etc.
	// Parent of j > 1 is (j-2)/k + 1
	for i := 2; i <= n; i++ {
		parent := (i-2)/ansK + 1
		p[i] = parent
		d[i] = d[parent] + 1
		degree[parent]++
		currSum += int64(d[i])
	}

	rem := s - currSum

	// Identify the "spine" to extend.
	// We choose the path from node n to 1 in the BFS tree as fixed.
	// Other nodes can be moved to the end of this spine to increase total depth.
	onPath := make([]bool, n+1)
	curr := n
	for curr != 0 {
		onPath[curr] = true
		curr = p[curr]
	}

	spineEnd := n

	// Movable nodes are those not on the fixed spine.
	// We process them in descending order of depth (BFS order n down to 2 essentially sorts by depth).
	movable := make([]int, 0, n)
	for i := n; i >= 2; i-- {
		if !onPath[i] {
			movable = append(movable, i)
		}
	}

	// Greedily move nodes to the end of the spine
	for _, u := range movable {
		if rem == 0 {
			break
		}

		// Calculate gain if we move u to be a child of current spineEnd
		gain := int64(d[spineEnd] + 1 - d[u])

		if rem >= gain {
			// Perform full move
			degree[p[u]]--
			p[u] = spineEnd
			degree[spineEnd]++
			d[u] = d[spineEnd] + 1
			spineEnd = u
			rem -= gain
		} else {
			// Perform partial move to satisfy exact rem
			// We need a target depth of d[u] + rem
			// So we need a parent at depth d[u] + rem - 1
			targetPDepth := d[u] + int(rem) - 1
			
			// Find a suitable parent with capacity
			found := false
			// We scan all nodes to find a valid parent.
			// Since we need to output any valid tree, finding the first one works.
			for v := 1; v <= n; v++ {
				if d[v] == targetPDepth && degree[v] < ansK {
					degree[p[u]]--
					p[u] = v
					degree[v]++
					d[u] = d[v] + 1
					rem = 0
					found = true
					break
				}
			}
			if !found {
				// This case should theoretically not be reachable given the constraints and logic
			}
			break
		}
	}

	// Output result
	for i := 2; i <= n; i++ {
		fmt.Fprintf(writer, "%d ", p[i])
	}
	fmt.Fprintln(writer)
}
```