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;
int n, m;
char s[2020];
long long f[2020][2020], ans, g[2020];
void U(long long &x, long long y) { x = (x + y) % 1000000007; }
int main() {
  cin >> n >> m >> s + 1;
  f[0][0] = 1;
  g[0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= m; ++j) {
      U(f[i][j], g[j] * (s[i] - 'a'));
      for (int k = i - 1; ~k && j + (i - k) * (n + 1 - i) <= m; --k)
        U(f[i][j + (i - k) * (n + 1 - i)], f[k][j] * ('z' - s[i]));
      U(g[j], f[i][j]);
    }
  }
  cout << g[m] << endl;
}
