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, ar[2200];
long long dn[2200][2200];
const long long mod = 1e9 + 7;
char s[2200];
int main() {
  cin >> N >> M;
  for (int i = 1; i <= N; i++) scanf(" %c", &s[i]);
  dn[0][0] = 1;
  for (int sum = 0; sum <= M; sum++) {
    long long k = (sum == 0);
    for (int i = 1; i <= N; i++) {
      for (int len = 0; i - len && sum >= (N - i + 1) * (len + 1); len++)
        dn[i][sum] =
            (dn[i][sum] +
             dn[i - len - 1][sum - (N - i + 1) * (len + 1)] * ('z' - s[i])) %
            mod;
      dn[i][sum] = (dn[i][sum] + k * (s[i] - 'a')) % mod;
      k = (k + dn[i][sum]);
    }
  }
  long long res = 0;
  for (int i = 0; i <= N; i++) res = (res + dn[i][M]) % mod;
  cout << res << endl;
  return 0;
}
