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
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:36777216")
using namespace std;
int n, m;
char ch[2222];
long long f[2222][2222], sum[2222][2222];
int main() {
  cin >> n >> m;
  scanf(" %s", ch + 1);
  f[0][0] = 1;
  for (int i = 0; i <= n; i++) sum[i][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= m; j++) {
      f[i][j] += sum[i - 1][j] * (ch[i] - 'a');
      f[i][j] %= 1000000007;
      for (int k = i - 1; k >= 0; k--) {
        if (j - (i - k) * (n - i + 1) < 0) break;
        f[i][j] += f[k][j - (i - k) * (n - i + 1)] * ('z' - ch[i]);
        f[i][j] %= 1000000007;
      }
      sum[i][j] = sum[i - 1][j] + f[i][j];
      sum[i][j] %= 1000000007;
    }
  }
  long long ans = 0;
  for (int i = 0; i <= n; i++) ans += f[i][m], ans %= 1000000007;
  printf("%I64d\n", ans);
}
