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 read() {
  long long x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
void writeln(long long x) {
  write(x);
  puts("");
}
const long long N = 2010, mod = 1e9 + 7;
long long los[N], win[N], n, K, f[N][N], g[N][N];
char s[N];
void add(long long &x, long long y) { x = (x + y) % mod; }
int main() {
  n = read();
  K = read();
  scanf("%s", s + 1);
  f[0][0] = g[0][0] = 1;
  for (long long i = 1; i <= n; ++i) los[i] = s[i] - 'a', win[i] = 'z' - s[i];
  los[n]++;
  for (long long i = 1; i <= n; ++i)
    for (long long k = 0; k <= K; ++k) {
      for (long long j = i - 1; (k >= (i - j) * (n - i + 1)) && (j >= 0); --j)
        add(f[i][k], f[j][k - (i - j) * (n - i + 1)] * win[i]);
      add(f[i][k], g[i - 1][k] * los[i]);
      g[i][k] = (f[i][k] + g[i - 1][k]) % mod;
    }
  writeln(f[n][K]);
}
