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 = 1e9 + 7;
int dp[2010][2010], sum[2010];
int add(int a, int b) {
  if (a + b >= mod) return a + b - mod;
  return a + b;
}
int mul(int a, int b) { return (long long)a * b % mod; }
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a % mod;
    }
    b >>= 1;
    a = a * a % mod;
  }
  return res;
}
char s[2010];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = (1); i < (n + 1); i++) cin >> s[i];
  dp[0][0] = sum[0] = 1;
  for (int i = (1); i < (n + 1); i++) {
    for (int j = (0); j < (k + 1); j++) {
      dp[i][j] = mul(s[i] - 'a', sum[j]);
      for (int q = 0; j - (q + 1) * (n - i + 1) >= 0 && q < i; q++) {
        dp[i][j] =
            add(mul('z' - s[i], dp[i - q - 1][j - (q + 1) * (n - i + 1)]),
                dp[i][j]);
      }
      sum[j] = add(sum[j], dp[i][j]);
    }
  }
  cout << sum[k] << "\n";
  return 0;
}
