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 long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long exp(long long a, long long b, long long m) {
  if (b == 0LL) return 1LL;
  if (b == 1LL) return mod(a, m);
  long long k = mod(exp(a, b / 2, m), m);
  if (b & 1LL) {
    return mod(a * mod(k * k, m), m);
  } else
    return mod(k * k, m);
}
long long n;
const long long N = 2020, M = 1e9 + 7;
long long dp[N][N][2];
char s[N];
long long solve(long long id, long long k, long long f) {
  if (k < 0) return 0;
  if (id == 0) return k == 0;
  long long& x = dp[id][k][f];
  if (x != -1) return x;
  x = 0;
  long long maior = 0;
  long long sefude = n - id + 1;
  for (long long put = 1; put <= id; put++) {
    if (put * sefude > k) break;
    maior =
        (maior + solve(id - put, k - put * sefude, 1) < M
             ? maior + solve(id - put, k - put * sefude, 1)
             : maior + solve(id - put, k - put * sefude, 1) -
                   (maior + solve(id - put, k - put * sefude, 1) - M) / M * M);
  }
  long long cur = s[id] - 'a';
  long long men = cur;
  if (men > 0)
    x = (x + cur * solve(id - 1, k, 0) < M
             ? x + cur * solve(id - 1, k, 0)
             : x + cur * solve(id - 1, k, 0) -
                   (x + cur * solve(id - 1, k, 0) - M) / M * M);
  if (!f)
    x = (x + solve(id - 1, k, 0) < M
             ? x + solve(id - 1, k, 0)
             : x + solve(id - 1, k, 0) - (x + solve(id - 1, k, 0) - M) / M * M);
  long long mai = 26 - cur - 1;
  x = (x + mai * maior < M ? x + mai * maior
                           : x + mai * maior - (x + mai * maior - M) / M * M);
  return x;
}
long long k;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  memset(dp, -1, sizeof(dp));
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> s[i];
  }
  long long res = solve(n, k, 0);
  res = mod(res, M);
  cout << res << "\n";
}
