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;
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  vector<vector<int> > dp(n + 1, vector<int>(k + 1));
  vector<int> sum(k + 1);
  dp[0][0] = 1;
  for (int i = 0; i < n; ++i) {
    const long long lo = (s[i] - 'a'), hi = 25 - lo;
    for (int j = 0; j <= k; ++j) {
      sum[j] = (sum[j] + dp[i][j]) % MOD;
    }
    const int m = n - i;
    for (int j = 0; j <= k; ++j) {
      int s = 0;
      for (int y = i, x = j - m; y >= 0 && x >= 0; --y, x -= m) {
        s = (s + dp[y][x]) % MOD;
      }
      dp[i + 1][j] = (dp[i + 1][j] + s * hi) % MOD;
      dp[i + 1][j] = (dp[i + 1][j] + sum[j] * lo) % MOD;
    }
  }
  int answer = 0;
  for (int i = 0; i <= n; ++i) {
    answer = (answer + dp[i][k]) % MOD;
  }
  cout << answer << endl;
  return 0;
}
