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;
long long dp[2050][2050], f[2050];
char s[2050];
int main() {
  int T, i, j, ca = 0;
  while (cin >> n >> k) {
    scanf("%s", s);
    memset(dp, 0, sizeof(dp));
    memset(f, 0, sizeof(f));
    dp[0][0] = 1;
    f[0] = 1;
    for (i = 1; i <= n; i++)
      for (j = 0; j <= k; j++) {
        dp[i][j] = (dp[i][j] + f[j] * (s[i - 1] - 'a')) % 1000000007;
        int c = i - 1;
        while (c >= 0 && (i - c) * (n - i + 1) <= j) {
          long long h = (i - c) * (n - i + 1);
          dp[i][j] = (dp[i][j] + dp[c][j - h] * ('z' - s[i - 1])) % 1000000007;
          c--;
        }
        f[j] = (f[j] + dp[i][j]) % 1000000007;
      }
    printf("%I64d\n", f[k]);
  }
  return 0;
}
