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 = 1e6;
int n, k, ans;
string s;
char check(char b, char c) {
  for (int i = 0; i < 26; i++) {
    if (b != i + 65 && i + 65 != c) return i + 65;
  }
}
int main() {
  scanf("%d%d", &n, &k);
  cin >> s;
  if (k > 2) {
    for (int i = 0; i < n - 1; i++) {
      if (s[i] == s[i + 1]) {
        if (i + 2 < n) {
          char x = check(s[i + 2], s[i + 1]);
          s[i + 1] = x;
        } else
          s[i + 1] = check('!', s[i + 1]);
        ans++;
      }
    }
    cout << ans << endl;
    cout << s << endl;
    return 0;
  } else {
    string t = s;
    int ans2 = 0;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) {
        if (s[i] != 'A') {
          s[i] = 'A';
          ans++;
        }
      } else {
        if (s[i] != 'B') {
          s[i] = 'B';
          ans++;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      if (i % 2 != 0) {
        if (t[i] != 'A') {
          t[i] = 'A';
          ans2++;
        }
      } else {
        if (t[i] != 'B') {
          t[i] = 'B';
          ans2++;
        }
      }
    }
    if (ans <= ans2) {
      cout << ans << endl;
      cout << s << endl;
    } else {
      cout << ans2 << endl;
      cout << t << endl;
    }
  }
  return 0;
}