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 long long mod = 1000000007;
int n, k;
long long dp[2005][2005][2];
string s;
int main() {
  cin >> n >> k >> s;
  dp[n][0][0] = 1;
  dp[n][0][1] = 1;
  for (int i = n - 1; i >= 0; --i) {
    for (int j = 0; j <= k; ++j) {
      for (int eq = 0; eq <= 1; ++eq) {
        dp[i][j][eq] = (dp[i][j][eq] + dp[i + 1][j][1]) % mod;
        dp[i][j][eq] =
            (dp[i][j][eq] + (dp[i + 1][j][0] * (s[i] - 'a') % mod)) % mod;
        if (!eq) {
          int till = -1;
          for (int a = 0; i + a < n; ++a) {
            if (j >= ((a + 1) * (n - (i + a)))) {
              dp[i][j][eq] += dp[i + a + 1][j - ((a + 1) * (n - (i + a)))][0] *
                              ('z' - s[a + i]);
              dp[i][j][eq] %= mod;
              till = a + 1;
            } else {
              break;
            }
          }
          if (till == -1) till = 0;
          for (int a = n - 1 - i; a >= till; --a) {
            if (j >= ((a + 1) * (n - (i + a)))) {
              dp[i][j][eq] += dp[i + a + 1][j - ((a + 1) * (n - (i + a)))][0] *
                              ('z' - s[a + i]);
              dp[i][j][eq] %= mod;
            } else
              break;
          }
        }
      }
    }
  }
  cout << dp[0][k][0] << "\n";
  return 0;
}
