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 int lower[2001][2001] = {0};
long long int great[2001][2001] = {0};
long long int qual[2001][2001] = {0};
long long int clower[2001] = {0};
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  int a = s.size() - 1;
  great[0][0] = 1;
  qual[1][0] = 1;
  great[1][1] = 'z' - s[a];
  lower[1][0] = 25 - ('z' - s[a]);
  clower[0] += lower[1][0];
  for (int i = 2; i <= s.size(); ++i) {
    for (int j = i; j <= (i * (i + 1)) / 2 && j <= 2000; ++j) {
      great[i][j] =
          (('z' - s[s.size() - i]) *
           (great[i - 1][j - i] + lower[i - 1][j - i] + qual[i - 1][j - i])) %
          1000000007;
    }
    for (int j = 0; j <= (i * (i + 1)) / 2 && j <= 2000; ++j) {
      qual[i][j] += clower[j];
      qual[i][j] = qual[i][j] % 1000000007;
    }
    for (int j = 0; j <= i / 2; ++j) {
      int f = (i - j) * j;
      for (int k = (i - j) * j;
           k - f <= max((j * (j + 1)) / 2, ((i - j) * (i - j + 1)) / 2) &&
           k <= 2000;
           ++k) {
        qual[i][k] += great[j][k - f];
        if (i - j != j && j != 0) qual[i][k] += great[i - j][k - f];
        qual[i][k] = qual[i][k] % 1000000007;
      }
    }
    for (int j = 0; j <= (i * (i + 1)) / 2 && j <= 2000; ++j) {
      lower[i][j] = ((25 - 'z' + s[s.size() - i]) *
                     (great[i - 1][j] + lower[i - 1][j] + qual[i - 1][j])) %
                    1000000007;
      clower[j] += lower[i][j];
      clower[j] = clower[j] % 1000000007;
    }
  }
  cout << (lower[n][m] + qual[n][m] + great[n][m]) % 1000000007 << endl;
  return 0;
}
