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 MAX = 5e5 + 9;
char s[MAX], a[MAX], b[MAX];
int main() {
  int n, k, ans = 0, c1, c2;
  scanf("%d%d%s", &n, &k, &s);
  if (k == 2) {
    c1 = c2 = 0;
    for (int i = 0; i < n; i++)
      if (i % 2) {
        a[i] = 'A', c1 += (s[i] == 'B');
        b[i] = 'B', c2 += (s[i] == 'A');
      } else {
        a[i] = 'B', c1 += (s[i] == 'A');
        b[i] = 'A', c2 += (s[i] == 'B');
      }
    return !printf("%d\n%s", min(c1, c2), c1 < c2 ? a : b);
  }
  for (int i = 1; s[i]; i++)
    if (s[i] == s[i - 1]) {
      for (int j = 0; j < k; j++)
        if (s[i - 1] != (j + 'A') && s[i + 1] != (j + 'A')) {
          s[i] = j + 'A';
          break;
        }
      ans++;
    }
  printf("%d\n%s", ans, s);
}