For problem statement at 1000-1999/1800-1899/1860-1869/1861/problemE.txt this is a correct solution, but verifier at 1000-1999/1800-1899/1860-1869/1861/verifierE.go ends with Runtime error on test 9: exit status 2

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

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

func main() {
	// Use buffered input/output for speed, though N=4000 is small enough for basic IO.
	in := bufio.NewReader(os.Stdin)
	var n, k int
	fmt.Fscan(in, &n, &k)

	// If k > n, no subarray of length k can be chosen. Cost is 0 for all arrays.
	if k > n {
		fmt.Println(0)
		return
	}

	const mod int64 = 998244353

	// Precompute powers of k modulo mod
	powK := make([]int64, n+1)
	powK[0] = 1
	k64 := int64(k)
	for i := 1; i <= n; i++ {
		powK[i] = (powK[i-1] * k64) % mod
	}

	// h[m] will store the number of arrays of length m such that:
	// 1. The array contains no valid subarray of length k (where valid means a permutation of 1..k).
	// 2. The suffix of length k-1 consists of distinct elements.
	// Note: Indices 0 to n-1 are allocated. We only use indices from k-1 to n-1.
	h := make([]int64, n)

	// dp[j] stores the number of "bad" arrays (no valid subarray) of the current length
	// such that the longest distinct suffix has length j.
	// Since the arrays are "bad", they cannot have a distinct suffix of length k (which would be a valid perm).
	// So j ranges from 1 to k-1. We use a slice of size k and ignore index 0.
	dp := make([]int64, k)

	// Base case: arrays of length 1.
	// Any array of length 1 is "bad" (since k >= 2) and has a distinct suffix of length 1.
	// There are k choices for the single element.
	dp[1] = k64

	// If k=2, then length 1 arrays have a distinct suffix of length k-1=1.
	// So h[1] should be initialized.
	if k == 2 {
		h[1] = dp[1]
	}

	newDp := make([]int64, k)
	suffixSum := make([]int64, k+1)

	// DP to compute h[i] for i from 2 to n-1
	for i := 2; i < n; i++ {
		// Calculate suffix sums of the previous DP state
		// suffixSum[j] = sum(dp[j...k-1])
		var currentSum int64 = 0
		for j := k - 1; j >= 1; j-- {
			currentSum = (currentSum + dp[j])
			if currentSum >= mod {
				currentSum -= mod
			}
			suffixSum[j] = currentSum
		}

		// Calculate new DP values for length i
		for j := 1; j < k; j++ {
			// Transition 1: Extend a distinct suffix of length j-1 to j.
			// This is possible if we pick one of the (k - (j-1)) available distinct values.
			// Valid for j > 1.
			var term1 int64
			if j > 1 {
				term1 = dp[j-1] * int64(k-j+1)
				term1 %= mod
			}

			// Transition 2: Reset the distinct suffix length to j.
			// This happens if we append a value that is already in the suffix,
			// such that the new distinct suffix has length j.
			// This sums up contributions from all previous lengths >= j.
			term2 := suffixSum[j]

			val := term1 + term2
			if val >= mod {
				val -= mod
			}
			newDp[j] = val
		}

		// Update dp array
		for j := 1; j < k; j++ {
			dp[j] = newDp[j]
		}

		// If the current length i is at least k-1, we record h[i].
		// h[i] corresponds to the count of bad arrays ending with k-1 distinct elements.
		if i >= k-1 {
			h[i] = dp[k-1]
		}
	}

	// S[i] stores the sum of costs for all arrays of length i.
	S := make([]int64, n+1)
	// S[0]...S[k-1] are 0 because no valid subarrays can be formed.

	for i := k; i <= n; i++ {
		var sumCosts int64 = 0
		// We iterate over the possible start position of the *first* valid subarray.
		// If the first valid subarray is A[j...j+k-1], then:
		// 1. A[1...j+k-2] contains no valid subarray.
		// 2. A[j...j+k-1] is a permutation.
		// This implies A[1...j+k-1] satisfies the conditions counted by h[j+k-2] extended by 1 element
		// to complete the permutation. There is exactly 1 way to complete the permutation given the prefix.
		// So we use h[j+k-2] ways for the prefix part.
		// The remaining suffix has length i - (j+k-1). Let this be 'rem'.
		// The contribution is 1 (for the current subarray) + cost of remaining suffix.
		// Total ways for suffix is k^rem.
		
		// rem ranges from 0 to i-k.
		// Corresponding prefix length m = i - 1 - rem.
		// We sum h[m] * (k^rem + S[rem]).

		for rem := 0; rem <= i-k; rem++ {
			m := i - 1 - rem
			ways := h[m]
			
			costSuffix := powK[rem] + S[rem]
			if costSuffix >= mod {
				costSuffix -= mod
			}
			
			term := (ways * costSuffix) % mod
			sumCosts = (sumCosts + term)
			if sumCosts >= mod {
				sumCosts -= mod
			}
		}
		S[i] = sumCosts
	}

	fmt.Println(S[n])
}
```