For problem statement at 0-999/400-499/490-499/494/problemB.txt this is a correct solution, but verifier at 0-999/400-499/490-499/494/verifierB.go ends with All 100 tests passed can you fix the verifier? ```go
package main

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

const MOD = 1000000007

func main() {
	reader := bufio.NewReader(os.Stdin)

	var s, t string
	fmt.Fscan(reader, &s)
	fmt.Fscan(reader, &t)

	n := len(s)
	m := len(t)

	// Build KMP failure function for t
	fail := make([]int, m)
	fail[0] = 0
	j := 0
	for i := 1; i < m; i++ {
		for j > 0 && t[i] != t[j] {
			j = fail[j-1]
		}
		if t[i] == t[j] {
			j++
		}
		fail[i] = j
	}

	// Find all positions in s where t ends (0-indexed end position)
	// endPos[i] = true means t ends at position i in s
	endPos := make([]bool, n)
	j = 0
	for i := 0; i < n; i++ {
		for j > 0 && s[i] != t[j] {
			j = fail[j-1]
		}
		if s[i] == t[j] {
			j++
		}
		if j == m {
			endPos[i] = true
			j = fail[j-1]
		}
	}

	// For each position i in s (1-indexed), compute:
	// earliest[i] = earliest start position (1-indexed) of a substring ending at or before i that contains t
	// i.e., if t ends at position e (0-indexed), then a valid substring containing t
	// must start at or before e - m + 2 (1-indexed) and end at or after e+1 (1-indexed)

	// Let's think in 1-indexed terms for s.
	// If t ends at position e+1 (1-indexed, so e is 0-indexed index),
	// then the substring [a, b] contains t if a <= e+1-m+1 = e-m+2 and b >= e+1.

	// We need to count sequences of non-overlapping substrings each containing t.

	// dp[i] = number of ways to choose substrings from s[1..i] (1-indexed)
	// Let dp[i] = number of valid selections where the last substring ends at position i
	// Plus a prefix sum approach.

	// Actually let's define:
	// f[i] = number of ways to select k>=1 non-overlapping substrings from s[1..i]
	//         where the last one ends exactly at i
	// g[i] = sum of f[1..i] = total ways where last substring ends at or before i

	// For f[i]: we need to choose the last substring [a, b] with b = i, and it contains t.
	// The earliest a can start such that [a, i] contains t is:
	//   the latest end of t that is <= i, call it lastT[i] (1-indexed end position)
	//   then a can be from 1 to lastT[i] - m + 1
	// The previous substring must end at most a-1.
	// So f[i] = sum over a from 1 to maxA[i] of (1 + g[a-1])
	//         = maxA[i] + sum_{a=1}^{maxA[i]} g[a-1]
	//         = maxA[i] + sum_{j=0}^{maxA[i]-1} g[j]

	// where maxA[i] = lastT[i] - m + 1, and lastT[i] is the largest j<=i where t ends at j (1-indexed)

	// Compute lastT array (1-indexed)
	lastT := make([]int, n+1) // lastT[i] for i in 1..n
	last := -1
	for i := 0; i < n; i++ {
		if endPos[i] {
			last = i + 1 // 1-indexed
		}
		lastT[i+1] = last
	}

	// g[i] = total ways with last substring ending at or before i (1-indexed), g[0] = 0
	g := make([]int64, n+2)
	// prefG[i] = g[0] + g[1] + ... + g[i]
	prefG := make([]int64, n+2)

	g[0] = 0
	prefG[0] = 0

	for i := 1; i <= n; i++ {
		lt := lastT[i]
		var fi int64
		if lt == -1 {
			fi = 0
		} else {
			maxA := lt - m + 1
			if maxA >= 1 {
				// f[i] = maxA + prefG[maxA-1] - prefG[-1]
				// = maxA + sum_{j=0}^{maxA-1} g[j]
				// = maxA + prefG[maxA-1]  (since prefG[j] = g[0]+...+g[j])
				fi = (int64(maxA) + prefG[maxA-1]) % MOD
			} else {
				fi = 0
			}
		}
		g[i] = (g[i-1] + fi) % MOD
		prefG[i] = (prefG[i-1] + g[i]) % MOD
	}

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