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;
;
struct _dbg {
  template <class T>
  _dbg& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  };
};
_dbg _d;
long long f[2010][2010], g[2010][2010], N, K, M = 1e9 + 7;
char s[2010];
int main() {
  cin >> N >> K >> (s + 1);
  f[0][0] = g[0][0] = 1;
  for (int i = (1), _i = (N + 1); i < _i; ++i)
    for (int k = 0, _k = (K + 1); k < _k; ++k) {
      f[i][k] = g[i - 1][k] * (s[i] - 'a') % M;
      for (int p = 0; p < i && (p + 1) * (N - i + 1) <= k; ++p) {
        (f[i][k] +=
         f[i - p - 1][k - (p + 1) * (N - i + 1)] * ('z' - s[i]) % M) %= M;
      };
      g[i][k] = (g[i - 1][k] + f[i][k]) % M;
    };
  cout << g[N][K] << '\n';
};
