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;
const int MOD = 1000000007;
const int MAX_N = 2010;
const int MAX_K = 2010;
int n, k, sol;
long long dp[MAX_N][MAX_K], sum[MAX_N][MAX_K];
char str[MAX_N];
int main() {
  cin >> n >> k;
  cin >> (str + 1);
  dp[0][0] = 1;
  sol = dp[0][k];
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= k; ++j) {
      for (int step = 1; step <= i && j - step * (n - i + 1) >= 0; ++step) {
        dp[i][j] =
            (dp[i][j] + dp[i - step][j - step * (n - i + 1)] * ('z' - str[i])) %
            MOD;
      }
      sum[i][j] = (sum[i - 1][j] + dp[i - 1][j]) % MOD;
      dp[i][j] = (dp[i][j] + sum[i][j] * (str[i] - 'a')) % MOD;
    }
    sol = (sol + dp[i][k]) % MOD;
  }
  cout << sol << endl;
  return 0;
}
