You are given a string S of length n with each character being one of the first m lowercase English letters. 

Calculate how many different strings T of length n composed from the first m lowercase English letters exist such that the length of LCS (longest common subsequence) between S and T is n - 1.

Recall that LCS of two strings S and T is the longest string C such that C both in S and T as a subsequence.

Input

The first line contains two numbers n and m denoting the length of string S and number of first English lowercase characters forming the character set for strings (1 ≤ n ≤ 100 000, 2 ≤ m ≤ 26).

The second line contains string S.

Output

Print the only line containing the answer.

Examples

Input

3 3
aaa


Output

6


Input

3 3
aab


Output

11


Input

1 2
a


Output

1


Input

10 9
abacadefgh


Output

789

Note

For the first sample, the 6 possible strings T are: aab, aac, aba, aca, baa, caa. 

For the second sample, the 11 possible strings T are: aaa, aac, aba, abb, abc, aca, acb, baa, bab, caa, cab.

For the third sample, the only possible string T is b.
#include <bits/stdc++.h>
using namespace std;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
const long double PI = 2 * acos(0.0);
const long double eps = 1e-15;
const int infi = 1e9;
const long long Linfi = 1e18;
const long long MOD = 1000000007;
const int c1 = 31;
const int c2 = 37;
long long n, m;
string s;
void solve() {
  cin >> n >> m >> s;
  m--;
  long long p = 0, c = 0;
  for (int i = 1; i < n; i++) {
    if (s[i] == s[i - 1]) {
      c += n * m;
      p = i;
    } else if (s[i] != s[i - 2])
      p = i - 1;
    c += i - p;
  }
  cout << n * n * m - c;
}
int main() {
  solve();
  return 0;
}
