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 powmod(long long int a, int b, int n) {
  long long int rm = 1;
  while (b) {
    if (b % 2) {
      rm = (rm * a) % n;
    }
    a = (a * a) % n;
    b /= 2;
  }
  return rm;
}
long long int dpl[2010][2010] = {0}, ctl, ct, dpg[2010][2010] = {0},
              prvdp[2010] = {0}, dp[2010][2010] = {0};
string s;
int n, k, mxreach;
int main() {
  cin >> n >> k >> s;
  prvdp[0] = dpl[n][0] = (s[n - 1] - 'a');
  dpg[n][1] = ('z' - s[n - 1]);
  dp[n][0] = s[n - 1] - 'a' + 1;
  dp[n][1] = 'z' - s[n - 1];
  for (int i = n - 1; i >= 1; i--) {
    for (int j = 0; j <= k; ++j) {
      mxreach = i + 1;
      for (int w = i + 1; w <= n; ++w) {
        if ((n - w + 1) * (w - i) > j) break;
        dp[i][j] += dpg[w][j - ((n - w + 1) * (w - i))];
        dp[i][j] %= int(1e9 + 7);
        mxreach = w;
      }
      for (int w = n; w > mxreach; --w) {
        if ((n - w + 1) * (w - i) > j) break;
        dp[i][j] += dpg[w][j - ((n - w + 1) * (w - i))];
        dp[i][j] %= int(1e9 + 7);
      }
      ctl = 'z' - s[i - 1];
      if (j >= n - i + 1) {
        dpg[i][j] += (ctl * (dp[i + 1][j - (n - i + 1)])) % int(1e9 + 7);
        dpg[i][j] %= int(1e9 + 7);
        dp[i][j] += dpg[i][j];
        dp[i][j] %= int(1e9 + 7);
      }
      ctl = s[i - 1] - 'a';
      dpl[i][j] =
          ((ctl * dp[i + 1][j]) % int(1e9 + 7) + prvdp[j]) % int(1e9 + 7);
      dp[i][j] += dpl[i][j];
      if (!j) dp[i][j]++;
      dp[i][j] %= int(1e9 + 7);
      prvdp[j] += (ctl * dp[i + 1][j]) % int(1e9 + 7);
      prvdp[j] %= int(1e9 + 7);
    }
  }
  cout << (dp[1][k] + int(1e9 + 7)) % int(1e9 + 7) << endl;
  return 0;
}
