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;
#pragma comment(linker, "/STACK:102400000,102400000")
const int MOD = 1e9 + 7;
const int N = 2007;
int n, k;
char s[N];
long long f[N][N], g[N][N];
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", s + 1);
  f[0][0] = 1;
  g[0][0] = 1;
  for (int i = int(1); i <= int(n); ++i) {
    for (int j = int(0); j <= int(k); ++j) {
      f[i][j] = g[i - 1][j] * (s[i] - 'a') % MOD;
      f[i][j] %= MOD;
      for (int l = 1; l <= i && l * (n + 1 - i) <= j; ++l) {
        f[i][j] += f[i - l][j - l * (n + 1 - i)] * ('z' - s[i]) % MOD;
        f[i][j] %= MOD;
      }
      g[i][j] = (g[i - 1][j] + f[i][j]) % MOD;
    }
  }
  cout << g[n][k] << endl;
  return 0;
}
