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;
long long dp[2050][2050];
long long p[2050][2050];
const int mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  s = " " + s;
  dp[0][0] = 1;
  p[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int sum = 0; sum <= k; sum++) {
      dp[i][sum] = (dp[i][sum] + p[i - 1][sum] * (s[i] - 'a')) % mod;
      for (int pref = 0; pref < i; pref++) {
        if (sum - (1 + pref) * (n - i + 1) >= 0) {
          dp[i][sum] =
              (dp[i][sum] + dp[i - pref - 1][sum - (1 + pref) * (n - i + 1)] *
                                ('z' - s[i])) %
              mod;
        } else {
          break;
        }
      }
    }
    for (int sum = 0; sum <= k; sum++) {
      p[i][sum] = (dp[i][sum] + p[i - 1][sum]) % mod;
    }
  }
  cout << p[n][k];
}
