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;
char s[2002];
int M[2002][2002];
int n, k;
long long solve(void) {
  for (long long score = 0; score <= k; ++score) {
    M[score][0] = 0;
    if (score == 0) {
      M[score][0] = s[0] - 'a';
    } else if (score == n) {
      M[score][0] = 'z' - s[0];
    }
    long long ss = M[score][0];
    for (long long i = 1; i < n; ++i) {
      long long total = 0;
      if (score == (i + 1) * (n - i)) {
        total = 'z' - s[i];
      } else if (score == 0) {
        total = s[i] - 'a';
      }
      total = (ss * (s[i] - 'a') + total) % 1000000007LL;
      for (long long j = i - 1; (j >= 0) && (score - (i - j) * (n - i) >= 0);
           --j) {
        total = (total +
                 M[score - (i - j) * (n - i)][j] * (long long)('z' - s[i])) %
                1000000007LL;
      }
      M[score][i] = total;
      ss = (ss + M[score][i]) % 1000000007LL;
    }
  }
  int sol = (k == 0);
  for (int i = 0; i < n; ++i) {
    sol = (sol + M[k][i]) % 1000000007LL;
  }
  return sol;
}
int main(int argc, char *argv[]) {
  cin >> n >> k >> s;
  int sol = solve();
  cout << sol << endl;
  return 0;
}
