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 nmax = (long long)1e6 + 10;
inline long long up(char ch) { return 'z' - ch; }
inline long long lo(char ch) { return ch - 'a'; }
inline long long eq(char ch) { return 1; }
long long dp[2002][2002];
long long tmp[2002];
signed main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  memset(dp, 0, sizeof dp);
  dp[n][0] = 1;
  for (long long i = n - 1; i >= 0; i--) {
    for (long long j = 0; j <= k; j++) {
      if (j - (n - i) >= 0)
        dp[i][j] = (up(s[i]) * dp[i + 1][j - (n - i)]) % 1000000007;
      dp[i][j] += (lo(s[i]) * dp[i + 1][j]) % 1000000007;
      dp[i][j] %= 1000000007;
      long long last = i;
      for (long long t = i + 1; t < n; t++) {
        if (j - (n - t) * (t - i + 1) >= 0) {
          dp[i][j] +=
              (up(s[t]) * dp[t + 1][j - (n - t) * (t - i + 1)]) % 1000000007;
          dp[i][j] %= 1000000007;
          last = t;
        } else {
          break;
        }
      }
      for (long long t = n - 1; t > last; t--) {
        if (j - (n - t) * (t - i + 1) >= 0) {
          dp[i][j] +=
              (up(s[t]) * dp[t + 1][j - (n - t) * (t - i + 1)]) % 1000000007;
          dp[i][j] %= 1000000007;
        } else {
          break;
        }
      }
      dp[i][j] += tmp[j];
      dp[i][j] %= 1000000007;
      if (j == 0) {
        dp[i][j]++;
        dp[i][j] %= 1000000007;
      }
      tmp[j] = tmp[j] + (lo(s[i]) * dp[i + 1][j]) % 1000000007;
      tmp[j] %= 1000000007;
    }
  }
  cout << dp[0][k] << endl;
  return 0;
}
