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>
const int MOD = 1e9 + 7;
const int ms = 2020;
int dp[ms][ms];
int eq[ms];
int main() {
  int n, k;
  std::cin >> n >> k;
  std::string str;
  std::cin >> str;
  str = '#' + str;
  n++;
  dp[0][0] = 1;
  for (int i = 0; i < n; i++) {
    int diff = 0;
    for (int sum = 0; sum <= k; sum++) {
      int v = dp[i][sum];
      dp[i][sum] = (dp[i][sum] + eq[sum] * (long long)(str[i] - 'a')) % MOD;
      eq[sum] = (eq[sum] + dp[i][sum]) % MOD;
    }
    long long choices = 1;
    for (int j = i + 1; j < n; j++) {
      diff = (j - i) * (n - j);
      for (int sum = 0; sum + diff <= k; sum++) {
        dp[j][sum + diff] =
            (dp[j][sum + diff] + (long long)dp[i][sum] * ('z' - str[j])) % MOD;
      }
    }
  }
  std::cout << (eq[k]) % MOD << std::endl;
}
