Problem: A colored stripe is represented by a horizontal row of n square cells, each cell is pained one of k colors. Your task is to repaint the minimum number of cells so that no two neighbouring cells are of the same color. You can use any color from 1 to k to repaint the cells.

Input

The first input line contains two integers n and k (1 ≤ n ≤ 5·105; 2 ≤ k ≤ 26). The second line contains n uppercase English letters. Letter "A" stands for the first color, letter "B" stands for the second color and so on. The first k English letters may be used. Each letter represents the color of the corresponding cell of the stripe.

Output

Print a single integer — the required minimum number of repaintings. In the second line print any possible variant of the repainted stripe.

Examples

Input

6 3
ABBACC


Output

2
ABCACA


Input

3 2
BBB


Output

1
BAB

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  long long int i, j, count = 0;
  if (k == 2) {
    string str1, str2;
    long long int cnt1 = 0, cnt2 = 0;
    for (i = 0; i < n; i++) {
      str1.push_back('A' + (i % 2));
      str2.push_back('A' + (i + 1) % 2);
    }
    for (i = 0; i < n; i++) {
      if (s[i] != str1[i]) cnt1++;
      if (s[i] != str2[i]) cnt2++;
    }
    if (cnt1 < cnt2)
      cout << cnt1 << endl << str1 << endl;
    else
      cout << cnt2 << endl << str2 << endl;
    return;
  }
  for (i = 1; i < n; i++) {
    if (s[i] == s[i - 1]) {
      for (j = 0; j < 3; j++) {
        if (s[i - 1] != 'A' + j && s[i + 1] != 'A' + j) break;
      }
      s[i] = 'A' + j;
      count++;
    }
  }
  cout << count << endl << s << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t = 1;
  while (t--) {
    solve();
  }
}