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 inf = 1e9 + 7;
template <class T>
void dbs(string str, T t) {
  cerr << str << " : " << t << "\n";
}
template <class T, class... second>
void dbs(string str, T t, second... s) {
  int idx = str.find(',');
  cerr << str.substr(0, idx) << " : " << t << ", ";
  dbs(str.substr(idx + 1), s...);
}
template <class second, class T>
ostream& operator<<(ostream& os, const pair<second, T>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <class T>
void prc(T a, T b) {
  cerr << "[";
  for (T i = a; i != b; ++i) {
    if (i != a) cerr << ", ";
    cerr << *i;
  }
  cerr << "]\n";
}
const int N = 2020;
long long first[N][N];
long long G[N][N];
string s;
long long g(int n, int k);
long long f(int n, int k) {
  if (k < 0) return 0;
  if (n == 0) return k == 0;
  long long& ans = first[n][k];
  if (ans != -1) return ans;
  ans = 0;
  int ch = s[int(s.size()) - n] - 'a';
  ans = ch * f(n - 1, k) + (25 - ch) * f(n - 1, k - n) + g(n - 1, k) + (k == 0);
  int nk = k;
  for (int i = 1; i <= n - 1; ++i) {
    int ch = s[int(s.size()) - n + i] - 'a';
    nk = k - (i + 1) * (n - i);
    if (nk < 0) {
      for (int i = n - 1; i >= 1; --i) {
        int ch = s[int(s.size()) - n + i] - 'a';
        nk = k - (i + 1) * (n - i);
        if (nk < 0) break;
        ans += (25 - ch) * f(n - i - 1, nk);
        ans %= inf;
      }
      break;
    }
    ans += (25 - ch) * f(n - i - 1, nk);
    ans %= inf;
  }
  ans %= inf;
  return ans;
}
long long g(int n, int k) {
  if (k < 0) return 0;
  if (n == 0) return 0;
  long long& ans = G[n][k];
  if (ans != -1) return ans;
  int ch = s[int(s.size()) - n] - 'a';
  ans = ch * f(n - 1, k) + g(n - 1, k);
  ans %= inf;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  memset(first, -1, sizeof first);
  memset(G, -1, sizeof G);
  cin >> s;
  cout << f(n, k) << "\n";
  return 0;
}
