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 main() {
  long long int n, k;
  scanf("%I64d%I64d", &n, &k);
  char s[n + 5];
  scanf("%s", s);
  long long int cnt = 0, ct = 0;
  if (k == 2) {
    for (long long int i = 0; i < n; i++) {
      if ((i + 1) % 2 == 0 && s[i] == 'A') {
        cnt++;
      } else if ((i + 1) % 2 && s[i] == 'B') {
        cnt++;
      }
      if ((i + 1) % 2 == 0 && s[i] == 'B') {
        ct++;
      } else if ((i + 1) % 2 && s[i] == 'A') {
        ct++;
      }
    }
    if (cnt <= ct) {
      for (long long int i = 0; i < n; i++) {
        if ((i + 1) % 2) {
          s[i] = 'A';
        } else {
          s[i] = 'B';
        }
      }
    } else {
      for (long long int i = 0; i < n; i++) {
        if ((i + 1) % 2) {
          s[i] = 'B';
        } else {
          s[i] = 'A';
        }
      }
    }
    cnt = min(cnt, ct);
  } else {
    for (long long int i = 1; i < n; i++) {
      if (i == n - 1 && s[i] == s[i - 1]) {
        cnt++;
        for (long long int j = 0; j < k; j++) {
          if (65 + j != (int)s[i - 1]) {
            s[i] = 65 + j;
            break;
          }
        }
      } else if (s[i] == s[i - 1] && i < n - 1) {
        cnt++;
        for (long long int j = 0; j < k; j++) {
          if (65 + j != (int)s[i - 1] && 65 + j != (int)s[i + 1]) {
            s[i] = 65 + j;
            break;
          }
        }
      }
    }
  }
  printf("%I64d\n", cnt);
  printf("%s\n", s);
  return 0;
}