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;
char k[500100];
char a[500100] = "A";
char b[500100] = "B";
int main() {
  int n, c;
  while (scanf("%d %d", &n, &c) != EOF) {
    for (int i = 1; i < n; i++) b[i] = a[i - 1], a[i] = b[i - 1];
    b[n] = a[n] = 0;
    gets(k);
    gets(k);
    int ans = 0;
    if (c > 2) {
      for (int i = 1; i < n; i++) {
        if (k[i] == k[i - 1]) {
          ans++;
          for (int j = 0; j < c; j++)
            if (j + 'A' != k[i - 1] && j + 'A' != k[i + 1]) {
              k[i] = j + 'A';
              break;
            }
        }
      }
      printf("%d\n%s\n", ans, k);
    } else {
      int tmp[2] = {0, 0};
      ;
      for (int i = 0; i < n; i++)
        tmp[0] += (k[i] != a[i]), tmp[1] += (k[i] != b[i]);
      printf("%d\n%s\n", tmp[0] > tmp[1] ? tmp[1] : tmp[0],
             tmp[0] > tmp[1] ? b : a);
    }
  }
}