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;
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long long int LINF = (long long int)1e18;
long long int gcd(long long int a, long long int b) {
  long long int r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
long long int fpow(long long int n, long long int k, int p = MOD) {
  long long int r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
void addmod(long long int &a, long long int val, long long int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
void submod(long long int &a, long long int val, long long int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
int mult(long long int a, long long int b, long long int p = MOD) {
  return (long long int)a * b % p;
}
int inv(long long int a, long long int p = MOD) { return fpow(a, p - 2, p); }
long long int n, m;
long long int ret = 0;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int ans = 0;
  cin >> n >> m;
  string second;
  cin >> second;
  m--;
  int flg = 0;
  for (int i = 1; i < n; i++) {
    if (second[i] == second[i - 1]) {
      flg = i;
      ans += n * m;
    } else if (i != 1 && second[i] != second[i - 2])
      flg = i - 1;
    ans += i - flg;
  }
  cout << n * n * m - ans << endl;
  return 0;
}
