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 ll = long long;
const int C = 2001, M = 1000000007;
ll dp[C][C], equal[C][C];
int main() {
  std::string a;
  int k, ln, i, j, ij, x;
  std::cin >> ln >> k >> a;
  dp[0][0] = a[0] - 'a';
  equal[0][0] = 1;
  i = 0, j = 0;
  for (ij = 0, x = ln - i; i + ij < ln && j + (ij + 1) * x <= k; ij++, x--) {
    dp[i + ij][j + (ij + 1) * x] += ('z' - a[i + ij]);
    dp[i + ij][j + (ij + 1) * x] %= M;
  }
  if (i + ij < ln) {
    for (ij = ln - i - 1, x = 1; ij > 0 && j + (ij + 1) * x <= k; ij--, x++) {
      dp[i + ij][j + (ij + 1) * x] += ('z' - a[i + ij]);
      dp[i + ij][j + (ij + 1) * x] %= M;
    }
  }
  for (i = 1; i < ln; i++) {
    for (j = 0; j <= k; j++) {
      equal[i][j] = (dp[i - 1][j] + equal[i - 1][j]) % M;
      dp[i][j] =
          (dp[i][j] + (dp[i - 1][j] + equal[i - 1][j]) * (a[i] - 'a')) % M;
      for (ij = 0, x = ln - i; i + ij < ln && j + (ij + 1) * x <= k;
           ij++, x--) {
        dp[i + ij][j + (ij + 1) * x] += dp[i - 1][j] * ('z' - a[i + ij]);
        dp[i + ij][j + (ij + 1) * x] %= M;
      }
      if (i + ij < ln) {
        for (ij = ln - i - 1, x = 1; ij > 0 && j + (ij + 1) * x <= k;
             ij--, x++) {
          dp[i + ij][j + (ij + 1) * x] += dp[i - 1][j] * ('z' - a[i + ij]);
          dp[i + ij][j + (ij + 1) * x] %= M;
        }
      }
    }
  }
  printf("%lld\n", (dp[ln - 1][k] + equal[ln - 1][k]) % M);
  return 0;
}
