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>
using namespace std;
int n, m, C[4005], S[4005][4005], Sum[4005][4005], k, Mi;
char p;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> p;
    C[i] = p - 'a';
  }
  S[n + 1][0] = Sum[n + 1][0] = 1;
  for (int i = n; i >= 1; i--)
    for (int j = 0; j <= m; j++) {
      S[i][j] = ((long long)C[i] * S[i + 1][j] + Sum[i + 1][j]) % 1000000007;
      Sum[i][j] = ((long long)C[i] * S[i + 1][j] + Sum[i + 1][j]) % 1000000007;
      for (k = i; k <= n; k++) {
        if (j >= (k - i + 1) * (n + 1 - k))
          S[i][j] = (S[i][j] + (long long)(25 - C[k]) *
                                   S[k + 1][j - (k - i + 1) * (n + 1 - k)] %
                                   1000000007) %
                    1000000007;
        else
          break;
      }
      Mi = k;
      for (int k = n; k >= Mi; k--) {
        if (j >= (k - i + 1) * (n + 1 - k))
          S[i][j] = (S[i][j] + (long long)(25 - C[k]) *
                                   S[k + 1][j - (k - i + 1) * (n + 1 - k)] %
                                   1000000007) %
                    1000000007;
        else
          break;
      }
    }
  cout << S[1][m];
}
