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;
const int maxn = 2005;
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int mul(int a, int b) { return 1ll * a * b % mod; }
int f[maxn][maxn], g[maxn][maxn], N, K;
string str;
signed main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (fopen("A.INP", "r")) {
    freopen("A.INP", "r", stdin);
    freopen("A.OUT", "w", stdout);
  }
  if (fopen(""
            ".inp",
            "r")) {
    freopen(
        ""
        ".inp",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  }
  cin >> N >> K >> str;
  str = " " + str;
  f[0][0] = g[0][0] = 1;
  for (int i = 1; i <= N; ++i) {
    for (int j = 0; j <= K; ++j) {
      add(f[i][j], mul(g[i - 1][j], str[i] - 'a'));
      for (int k = i - 1; k >= 0 && j + (i - k) * (N + 1 - i) <= K; --k) {
        add(f[i][j + (i - k) * (N + 1 - i)], mul(f[k][j], 'z' - str[i]));
      }
      add(g[i][j], f[i][j]);
      add(g[i][j], g[i - 1][j]);
    }
  }
  cout << g[N][K] << '\n';
}
