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>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
int n, k;
long long dp[2001][2001], tot[2001];
string st;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k >> st;
  dp[0][0] = 1;
  tot[0] = 1;
  for (int i = 1; i < n + 1; i++)
    for (int j = 0; j < k + 1; j++) {
      dp[i][j] = (1ll * tot[j] * (st[i - 1] - 'a')) % 1000000007;
      for (int p = 0; p < i; p++) {
        int rem = j - ((p + 1) * (n - i + 1));
        if (rem < 0) break;
        dp[i][j] =
            ((dp[i][j] + dp[i - p - 1][rem] * ('z' - st[i - 1]))) % 1000000007;
      }
      tot[j] = (tot[j] + dp[i][j]) % 1000000007;
    }
  cout << tot[k];
  return 0;
}
