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 N = 105;
int main() {
  string s;
  int n, k, cnt = 0;
  scanf("%d%d", &n, &k);
  cin >> s;
  if (k == 2) {
    char x[] = {'A', 'B'};
    int q = 0, cnt1 = 0, cnt2 = 0;
    string temp1, temp2;
    for (int i = 0; i < n; i++, temp1 += x[q], q = (q + 1) & 1)
      if (s[i] != x[q]) {
        cnt1++;
      }
    char X[] = {'B', 'A'};
    q = 0;
    for (int i = 0; i < n; i++, temp2 += X[q], q = (q + 1) & 1)
      if (s[i] != X[q]) {
        cnt2++;
      }
    cout << min(cnt1, cnt2) << endl << ((cnt1 < cnt2) ? temp1 : temp2);
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (s[i] == s[i - 1]) {
      for (int j = 0; j < k; j++)
        if (s[i - 1] != char(j + 'A') &&
            (i + 1 >= n || s[i + 1] != char(j + 'A'))) {
          s[i] = char(j + 'A');
          cnt++;
          break;
        }
    }
  cout << cnt << endl << s;
  return 0;
}