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, k;
char C[3000];
long long D[3000][3000];
long long S[3000];
long long s = 0;
int main() {
  cin >> n >> k;
  cin >> C;
  S[0] = D[0][0] = (C[0] - 'a');
  S[n] = D[0][n] = ('z' - C[0]);
  s = D[0][k];
  for (int i = 1; i < n; i++) {
    int z = (i + 1) * (n - i);
    D[i][0] += (C[i] - 'a');
    if (z <= k) D[i][z] += ('z' - C[i]);
    for (int j = 0; j < i; j++) {
      int z = (i - j) * (n - i);
      for (int x = z; x <= k; x++) D[i][x] += ('z' - C[i]) * D[j][x - z];
    }
    for (int j = 0; j <= k; j++) {
      D[i][j] += (C[i] - 'a') * S[j];
      D[i][j] %= 1000000007;
      S[j] += D[i][j];
      S[j] %= 1000000007;
    }
    s += D[i][k];
    s %= 1000000007;
  }
  if (k == 0) s++;
  cout << s % 1000000007 << endl;
  return 0;
}
