#include <bits/stdc++.h>
int N, K, f[2005][2005], g[2005][2005];
char S[2005];
void doit() {
  scanf("%d%d%s", &N, &K, S + 1), f[0][0] = g[0][0] = 1;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j <= K; j++) {
      f[i][j] = 1ll * g[i - 1][j] * (S[i] - 97) % 1000000007;
      for (int k = 0, t; k < i && (t = (k + 1) * (N - i + 1)) <= j; k++)
        (f[i][j] += 1ll * f[i - k - 1][j - t] * (122 - S[i]) % 1000000007) %=
            1000000007;
      g[i][j] = (g[i - 1][j] + f[i][j]) % 1000000007;
    }
  printf("%d\n", g[N][K]);
}
int main() {
  doit();
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

Levko loves strings of length n, consisting of lowercase English letters, very much. He has one such string s. For each string t of length n, Levko defines its beauty relative to s as the number of pairs of indexes i, j (1 ≤ i ≤ j ≤ n), such that substring t[i..j] is lexicographically larger than substring s[i..j].

The boy wondered how many strings t are there, such that their beauty relative to s equals exactly k. Help him, find the remainder after division this number by 1000000007 (109 + 7).

A substring s[i..j] of string s = s1s2... sn is string sisi + 1... sj.

String x = x1x2... xp is lexicographically larger than string y = y1y2... yp, if there is such number r (r < p), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1. The string characters are compared by their ASCII codes.

Input

The first line contains two integers n and k (1 ≤ n ≤ 2000, 0 ≤ k ≤ 2000).

The second line contains a non-empty string s of length n. String s consists only of lowercase English letters. 

Output

Print a single number — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

2 2
yz


Output

26


Input

2 3
yx


Output

2


Input

4 7
abcd


Output

21962

### Response

```cpp
#include <bits/stdc++.h>
int N, K, f[2005][2005], g[2005][2005];
char S[2005];
void doit() {
  scanf("%d%d%s", &N, &K, S + 1), f[0][0] = g[0][0] = 1;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j <= K; j++) {
      f[i][j] = 1ll * g[i - 1][j] * (S[i] - 97) % 1000000007;
      for (int k = 0, t; k < i && (t = (k + 1) * (N - i + 1)) <= j; k++)
        (f[i][j] += 1ll * f[i - k - 1][j - t] * (122 - S[i]) % 1000000007) %=
            1000000007;
      g[i][j] = (g[i - 1][j] + f[i][j]) % 1000000007;
    }
  printf("%d\n", g[N][K]);
}
int main() {
  doit();
  return 0;
}

```