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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
const int N = 2005, mod = 1e9 + 7;
int dp[N][N], sum[N];
char s[N];
void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  int n = read(), k = read();
  scanf("%s", s + 1);
  sum[0] = dp[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= k; ++j) {
      for (int l = i - 1; l >= 0 && (i - l) * (n - i + 1) <= j; --l)
        add(dp[i][j], dp[l][j - (i - l) * (n - i + 1)]);
      dp[i][j] = 1ll * ('z' - s[i]) * dp[i][j] % mod;
      add(dp[i][j], 1ll * sum[j] * (s[i] - 'a') % mod);
      add(sum[j], dp[i][j]);
    }
  }
  int ans = 0;
  for (int i = 0; i <= n; ++i) add(ans, dp[i][k]);
  cout << ans;
  return 0;
}
