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 dp[2005][2005], MOD = 1e9 + 7, prefix[2005];
string S;
int n, k;
void add(long long &a, long long b) {
  a += b;
  while (a >= MOD) a -= MOD;
}
int main() {
  cin >> n >> k;
  cin >> S;
  S = '0' + S;
  dp[0][0] = 1;
  prefix[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      add(dp[i][j], (prefix[j] * (S[i] - 'a')) % MOD);
      for (int ii = i - 1; ii >= 0; ii--) {
        if (j - (i - ii) * (n - i + 1) >= 0)
          add(dp[i][j],
              (dp[ii][j - (i - ii) * (n - i + 1)] * ('z' - S[i])) % MOD);
        else
          break;
      }
      add(prefix[j], dp[i][j]);
    }
  }
  long long sum = 0;
  for (int i = 1; i <= n; i++) add(sum, dp[i][k]);
  if (k == 0) {
    add(sum, 1);
    cout << sum;
    return 0;
  }
  cout << sum;
}
