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 maxN = 3000;
const long long MOD = 1000 * 1000 * 1000 + 7;
long long n, k, dp[maxN][maxN], sum[maxN][maxN], s[maxN], ans;
char c;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> c;
    s[i] = c;
  }
  sum[0][0] = dp[0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= k; j++) {
      dp[i + 1][j] += sum[i][j] * (s[i] - 'a');
      dp[i + 1][j] %= MOD;
      for (int x = 0; i - x >= 0 && j >= (x + 1) * (n - i); x++)
        dp[i + 1][j] =
            (dp[i + 1][j] + dp[i - x][j - (x + 1) * (n - i)] * ('z' - s[i])) %
            MOD;
      sum[i + 1][j] += (sum[i][j] + dp[i + 1][j]) % MOD;
    }
  }
  for (int i = 0; i <= n; i++) ans = (ans + dp[i][k]) % MOD;
  cout << ans % MOD << endl;
}
