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;
long long d[3000][3000], t[3000][3000];
const long long MOD = 1000000007;
string s;
long long g(char c) { return 'z' - c; }
long long l(char c) { return c - 'a'; }
int main() {
  int n, k;
  cin >> n >> k;
  cin >> s;
  reverse(s.begin(), s.end());
  d[0][0] = 1;
  for (long long num = 1; num <= n; num++) {
    for (long long j = 0; j <= k; j++) {
      long long i;
      if (j == 0) {
        d[num][j] = 1;
      }
      t[num][j] = (t[num - 1][j] + (l(s[num - 1]) * d[num - 1][j])) % MOD;
      for (i = num - 1; i >= 0; i--) {
        long long r = (num - i) * (i + 1);
        if (r > j) {
          break;
        }
        d[num][j] += (d[i][j - r] * g(s[i])) % MOD;
        d[num][j] %= MOD;
      }
      for (long long it = 0; it < i; it++) {
        long long r = (num - it) * (it + 1);
        if (r > j) {
          break;
        }
        d[num][j] += (d[it][j - r] * g(s[it])) % MOD;
        d[num][j] %= MOD;
      }
      d[num][j] += t[num][j];
      d[num][j] %= MOD;
    }
  }
  cout << d[n][k];
  return 0;
}
