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;
const int mod = 1e9 + 7;
const int N = 2010;
char s[N];
int f[N][N], g[N][N];
int main() {
  int n, K;
  scanf("%d %d %s", &n, &K, s + 1);
  auto _add = [](int &x, int y) {
    x += y;
    if (x >= mod) x -= mod;
    return x;
  };
  f[0][0] = 1;
  g[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= K; j++) {
      for (int k = i - 1; k >= 0; k--) {
        if ((i - k) * (n - i + 1) > j) break;
        _add(f[i][j],
             ('z' - s[i]) * 1ll * f[k][j - ((i - k) * (n - i + 1))] % mod);
      }
      _add(f[i][j], (s[i] - 'a') * 1ll * g[i - 1][j] % mod);
      _add(g[i][j], g[i - 1][j]);
      _add(g[i][j], f[i][j]);
    }
  }
  int ans = 0;
  for (int i = 0; i <= n; i++) _add(ans, f[i][K]);
  cout << ans;
  return 0;
}
