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 maxn = 2017;
const int MOD = 1000000007;
int f[maxn][maxn], g[maxn][maxn], n, K, ans;
char s[maxn];
int main() {
  scanf("%d%d", &n, &K);
  scanf("%s", s + 1);
  f[0][0] = 1;
  g[0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= K; j++) {
      f[i][j] = 1LL * g[i - 1][j] * (s[i] - 'a') % MOD;
      for (int pre = 0; pre <= i - 1 && (pre + 1) * (n - i + 1) <= j; pre++)
        f[i][j] = (f[i][j] + 1LL * f[i - 1 - pre][j - (pre + 1) * (n - i + 1)] *
                                 ('z' - s[i])) %
                  MOD;
      g[i][j] = (g[i - 1][j] + f[i][j]) % MOD;
    }
  for (int i = 0; i <= n; i++) ans = (ans + f[i][K]) % MOD;
  printf("%d\n", ans);
  return 0;
}
