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 N = 2e3 + 10;
const int mod = 1e9 + 7;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
int n, m, f[N][N], g[N];
char s[N];
int main() {
  cin >> n >> m;
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) s[i] -= 'a';
  f[n + 1][0] = 1;
  for (int i = n; i; i--) {
    for (int k = 0; k <= m; k++) {
      g[k] = (g[k] + 1ll * s[i] * f[i + 1][k]) % mod;
      f[i][k] = g[k];
      for (int j = i;
           j <= n && j - i + 1 <= n - j + 1 && (j - i + 1) * (n - j + 1) <= k;
           j++)
        f[i][k] = (f[i][k] + 1ll * (25 - s[j]) *
                                 f[j + 1][k - (j - i + 1) * (n - j + 1)]) %
                  mod;
      for (int j = n;
           j >= i && j - i + 1 > n - j + 1 && (j - i + 1) * (n - j + 1) <= k;
           j--)
        f[i][k] = (f[i][k] + 1ll * (25 - s[j]) *
                                 f[j + 1][k - (j - i + 1) * (n - j + 1)]) %
                  mod;
    }
    f[i][0] = (f[i][0] + 1) % mod;
  }
  cout << f[1][m];
  return 0;
}
