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;
int main() {
  int n, k, ans = 0, check1 = 0, check2 = 0;
  string s;
  cin >> n >> k >> s;
  if (k == 2) {
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0 && s[i] == 'A') check1++;
      if (i % 2 == 1 && s[i] == 'B') check1++;
    }
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0 && s[i] == 'B') check2++;
      if (i % 2 == 1 && s[i] == 'A') check2++;
    }
    if (check1 <= check2) {
      cout << check1 << endl;
      for (int i = 0; i < n; i++) i % 2 == 0 ? cout << 'B' : cout << 'A';
    } else {
      cout << check2 << endl;
      for (int i = 0; i < n; i++) i % 2 == 0 ? cout << 'A' : cout << 'B';
    }
    return 0;
  } else {
    for (int i = 1; i < n; i++) {
      if (s[i] == s[i - 1]) {
        ans++;
        if (s[i - 1] != 'A' && s[i + 1] != 'A')
          s[i] = 'A';
        else if (s[i - 1] != 'B' && s[i + 1] != 'B')
          s[i] = 'B';
        else
          s[i] = 'C';
      }
    }
    cout << ans << endl << s;
  }
}