#include <bits/stdc++.h>
const int M = 2005;
long long f[M][M], g[M][M];
char S[M];
int n, m;
int main() {
  scanf("%d%d%s", &n, &m, S + 1);
  f[0][0] = g[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= m; ++j) {
      f[i][j] = g[i - 1][j] * (S[i] - 'a' + (i == n));
      int l = 1, r = i;
      while (l <= r && (i - l + 1) * (n - i + 1) <= j) ++l;
      while (l <= r && (i - r + 1) * (n - i + 1) <= j) --r;
      for (int k = 1; k < l; ++k)
        f[i][j] += f[k - 1][j - (i - k + 1) * (n - i + 1)] * ('z' - S[i]);
      for (int k = i; k > r; --k)
        f[i][j] += f[k - 1][j - (i - k + 1) * (n - i + 1)] * ('z' - S[i]);
      f[i][j] %= 1000000007;
    }
    for (int j = 0; j <= m; ++j) g[i][j] = (g[i - 1][j] + f[i][j]) % 1000000007;
  }
  printf("%lld\n", f[n][m]);
  return 0;
}

### Prompt

Please formulate a CPP solution to 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>
const int M = 2005;
long long f[M][M], g[M][M];
char S[M];
int n, m;
int main() {
  scanf("%d%d%s", &n, &m, S + 1);
  f[0][0] = g[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= m; ++j) {
      f[i][j] = g[i - 1][j] * (S[i] - 'a' + (i == n));
      int l = 1, r = i;
      while (l <= r && (i - l + 1) * (n - i + 1) <= j) ++l;
      while (l <= r && (i - r + 1) * (n - i + 1) <= j) --r;
      for (int k = 1; k < l; ++k)
        f[i][j] += f[k - 1][j - (i - k + 1) * (n - i + 1)] * ('z' - S[i]);
      for (int k = i; k > r; --k)
        f[i][j] += f[k - 1][j - (i - k + 1) * (n - i + 1)] * ('z' - S[i]);
      f[i][j] %= 1000000007;
    }
    for (int j = 0; j <= m; ++j) g[i][j] = (g[i - 1][j] + f[i][j]) % 1000000007;
  }
  printf("%lld\n", f[n][m]);
  return 0;
}

```