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>
#pragma comment(linker, "/STACK:100000000000,100000000000")
using namespace std;
const long long inf = 1e18 + 7;
const long long mod = 1e9 + 7;
const double eps = 1e-12;
const double PI = 2 * acos(0.0);
const double E = 2.71828;
int n, k;
char s[2005];
int f[2005][2005];
int tot[2005];
void add(int &x, int y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
}
int main(void) {
  scanf("%d%d\n", &n, &k);
  gets(s + 1);
  f[0][0] = 1;
  tot[0] = 1;
  s[0] = 'z';
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= k; ++j) {
      for (int q = 1; q <= i && j - q * (n - i + 1) >= 0; ++q) {
        add(f[i][j], f[i - q][j - q * (n - i + 1)] * 1LL * ('z' - s[i]) % mod);
      }
      add(f[i][j], tot[j] * 1LL * (s[i] - 'a') % mod);
    }
    for (int j = 0; j <= k; ++j) {
      add(tot[j], f[i][j]);
    }
  }
  cout << tot[k] << endl;
  return 0;
}
