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;
int n, k, ansm, ko, lo, ans;
string s, alph;
int main() {
  scanf("%d%d", &n, &k);
  cin >> s;
  if (k == 2) {
    for (int i = 0; i < s.size(); ++i) {
      if (i % 2 == 1 && s[i] == 'A') {
        ++ans;
      }
      if (i % 2 == 0 && s[i] == 'B') {
        ++ans;
      }
    }
    for (int i = 0; i < s.size(); ++i) {
      if (i % 2 == 1 && s[i] == 'B') {
        ++ko;
      }
      if (i % 2 == 0 && s[i] == 'A') {
        ++ko;
      }
    }
    if (ans <= ko) {
      for (int i = 0; i < s.size(); ++i) {
        if (i % 2 == 1 && s[i] == 'A') {
          s[i] = 'B';
        }
        if (i % 2 == 0 && s[i] == 'B') {
          s[i] = 'A';
        }
      }
    } else {
      for (int i = 0; i < s.size(); ++i) {
        if (i % 2 == 1 && s[i] == 'B') {
          s[i] = 'A';
        }
        if (i % 2 == 0 && s[i] == 'A') {
          s[i] = 'B';
        }
      }
    }
    cout << min(ans, ko) << endl;
    cout << s;
  } else {
    for (int i = 0; i < s.size(); ++i) {
      if (s[i] == s[i + 1]) {
        ++ans;
        char ansc = 'A';
        while (ansc == s[i] || ansc == s[i + 2]) {
          ansc++;
        }
        s[i + 1] = ansc;
      }
    }
    cout << ans << endl;
    cout << s;
  }
  return 0;
}