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;
long long dp[2003][2003], sp[2003][2003];
int main() {
  int n, k, i, j, big, l, tmp = 1;
  char s[2005];
  cin >> n >> k;
  cin >> s;
  dp[n][0] = sp[n][0] = 1;
  for (i = n - 1; i >= 0; i--) {
    dp[i][0] = ((dp[i + 1][0]) * (s[i] - 'a' + 1LL)) % (1000000007LL);
    sp[i][0] = dp[i][0];
    for (j = 1; j <= k; j++) {
      sp[i][j] = (dp[i + 1][j] * (s[i] - 'a') + sp[i + 1][j]) % (1000000007LL);
      dp[i][j] = sp[i][j];
      for (l = 0; (j >= ((n - i - l) * (l + 1))) && l < (n - i); l++) {
        dp[i][j] = (dp[i][j] + (25LL - s[i + l] + 'a') *
                                   dp[i + l + 1][j - (n - i - l) * (l + 1)]) %
                   (1000000007LL);
      }
      if (l < n - i) {
        for (l = n - i - 1; j >= ((n - i - l) * (l + 1)); l--) {
          dp[i][j] = (dp[i][j] + (25LL - s[i + l] + 'a') *
                                     dp[i + l + 1][j - (n - i - l) * (l + 1)]) %
                     (1000000007LL);
        }
      }
    }
  }
  cout << dp[0][k] << endl;
  return 0;
}
