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 N = 2010;
const long long Mod = 1000000007LL;
long long dp[N][N], sum[N];
char s[N];
int n, K;
void solve() {
  scanf("%s", s + 1);
  memset(dp, 0, sizeof(dp));
  memset(sum, 0, sizeof(sum));
  dp[0][0] = 1;
  sum[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int k = 0; k <= K; k++) {
      dp[i][k] = sum[k] * (s[i] - 'a') % Mod;
      for (int j = i - 1; j >= 0; j--) {
        int v = (i - j) * (n - i + 1);
        if (v > k) break;
        dp[i][k] = (dp[i][k] + dp[j][k - v] * ('z' - s[i]) % Mod) % Mod;
      }
      sum[k] = (sum[k] + dp[i][k]) % Mod;
    }
  }
  printf("%I64d\n", sum[K]);
}
int main() {
  while (scanf("%d%d", &n, &K) != EOF) solve();
  return 0;
}
