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;
const int maxn = 500000;
int n, k;
char str[maxn + 10];
int main() {
  cin >> n >> k;
  scanf("%s", str);
  int ans = 0;
  if (k == 2) {
    int A = 0, B = 0;
    for (int i = 0; str[i]; i++) {
      if (i & 1) {
        if (str[i] == 'A')
          B++;
        else
          A++;
      } else {
        if (str[i] == 'A')
          A++;
        else
          B++;
      }
    }
    if (A < B) {
      cout << A << endl;
      for (int i = 0; str[i]; i++) {
        if (i & 1)
          printf("A");
        else
          printf("B");
      }
      puts("");
    } else {
      cout << B << endl;
      for (int i = 0; str[i]; i++) {
        if (i & 1)
          printf("B");
        else
          printf("A");
      }
      puts("");
    }
  } else {
    for (int i = 1; str[i]; i++) {
      if (str[i] == str[i - 1]) {
        ans++;
        for (int j = 0; j < 3; j++) {
          if (str[i - 1] != 'A' + j) {
            if (str[i + 1]) {
              if (str[i + 1] != 'A' + j) {
                str[i] = 'A' + j;
                break;
              }
            } else {
              str[i] = 'A' + j;
              break;
            }
          }
        }
      }
    }
    cout << ans << endl;
    cout << str << endl;
  }
  return 0;
}