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 maxn = 1e5 + 200;
const int inf = 0x3f3f3f3f;
const double eps = 1e-7;
const double pi = acos(-1.0);
const int mod = 1e9 + 7;
inline int lowbit(int x) { return x & (-x); }
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void ex_gcd(long long a, long long b, long long &d, long long &x,
            long long &y) {
  if (!b) {
    d = a, x = 1, y = 0;
  } else {
    ex_gcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
inline long long qpow(long long a, long long b, long long MOD = mod) {
  long long res = 1;
  a %= MOD;
  while (b > 0) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return res;
}
inline long long inv(long long x, long long p) { return qpow(x, p - 2, p); }
inline long long Jos(long long n, long long k, long long s = 1) {
  long long res = 0;
  for (int i = 1; i <= n + 1; ++i) res = (res + k) % i;
  return (res + s) % n;
}
inline long long read() {
  long long f = 1;
  long long x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * f;
}
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
vector<char> block;
int main() {
  long long n = read(), m = read();
  string s;
  cin >> s;
  block.push_back(s[0]);
  for (int i = 1; i < n; i++) {
    if (s[i] != s[i - 1]) block.push_back(s[i]);
  }
  long long ans = n * (m - 1);
  for (int i = 1; i <= block.size() - 1; ++i) ans += n * m - n;
  long long len = 1;
  for (int i = 1; i <= n - 1; ++i) {
    if (len == 1 && s[i] != s[i - 1])
      len++;
    else if (len == 1)
      continue;
    else {
      if (s[i] == s[i - 2])
        len++;
      else {
        ans -= len * (len - 1) / 2;
        len = 1;
        if (s[i] != s[i - 1]) len = 2;
      }
    }
  }
  ans -= len * (len - 1) / 2;
  cout << ans << '\n';
  return 0;
}
