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;
int n, k, MD = 1e9 + 7, dp[2030][2040][5];
char bb[2030];
long long cum[2090][2090];
int solve(int x, int last, int f) {
  if (last < 0) return 0;
  if (x == n) return (int)(last == 0);
  int& ans = dp[x][last][f];
  if (ans != -1) return ans;
  ans = (1ll * (bb[x] - 'a') * solve(x + 1, last, 0)) % MD;
  int zbr = bb[x] - 'a';
  zbr = 26 - zbr;
  ans += solve(x + 1, last, 1);
  if (ans > MD) ans -= MD;
  if (f == 1) return ans;
  int j;
  for (j = x; j < n; j++) {
    if (last - (j - x + 1) * (n - j) < 0) break;
    ans +=
        (1ll * ('z' - bb[j]) * solve(j + 1, last - (j - x + 1) * (n - j), 0)) %
        MD;
    if (ans > MD) ans -= MD;
  }
  for (int i = n - 1; i > j; i--) {
    if (last - (i - x + 1) * (n - i) < 0) break;
    ans +=
        (1ll * ('z' - bb[i]) * solve(i + 1, last - (i - x + 1) * (n - i), 0)) %
        MD;
    if (ans > MD) ans -= MD;
  }
  return ans;
}
int main() {
  cin >> n >> k;
  memset(dp, -1, sizeof dp);
  scanf("%s", bb);
  for (int i = 0; i < n; i++) {
    long long zbr = 1;
    for (int j = i; j < n; j++) {
      cum[i][j] = zbr;
      zbr = (zbr * (26 - bb[j] + 'a')) % MD;
    }
    cum[i][n] = zbr;
  }
  cout << solve(0, k, 0);
}
