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;
  cin >> n >> k;
  string s;
  cin >> s;
  int ans = 0;
  if (k == 2) {
    int Ae = 0, Be = 0, Bo = 0, Ao = 0;
    for (int i = 0; i < s.size(); i++) {
      if (i % 2) {
        if (s[i] == 'A')
          Ao++;
        else
          Bo++;
      } else {
        if (s[i] == 'A')
          Ae++;
        else
          Be++;
      }
    }
    if (Ae + Bo < Ao + Be) {
      ans = Ae + Bo;
      for (int i = 0; i < s.size(); i++) {
        if (!(i % 2))
          s[i] = 'B';
        else
          s[i] = 'A';
      }
    } else {
      ans = Ao + Be;
      for (int i = 0; i < s.size(); i++) {
        if (!(i % 2))
          s[i] = 'A';
        else
          s[i] = 'B';
      }
    }
  } else {
    for (int i = 0; i < (int)s.size() - 1; i++)
      if (s[i] == s[i + 1]) {
        if (i + 2 < s.size()) {
          if (s[i + 1] == s[i + 2]) {
            ans++;
            if (s[i] == 'A')
              s[i + 1] = 'B';
            else
              s[i + 1] = 'A';
          } else {
            ans++;
            int a[3] = {0};
            if ((int)(s[i + 1] - 'A') < 3) a[(int)(s[i + 1] - 'A')] = 1;
            if (i - 1 >= 0 && (int)(s[i - 1] - 'A') < 3)
              a[(int)(s[i - 1] - 'A')] = 1;
            for (int j = 0; j < 3; j++)
              if (!a[j]) {
                s[i] = 'A' + j;
                break;
              }
          }
        } else {
          ans++;
          int a[3] = {0};
          if ((int)(s[i + 1] - 'A') < 3) a[(int)(s[i + 1] - 'A')] = 1;
          if (i - 1 >= 0) a[(int)(s[i - 1] - 'A')] = 1;
          for (int j = 0; j < 3; j++)
            if (!a[j]) {
              s[i] = 'A' + j;
              break;
            }
        }
      }
  }
  cout << ans << '\n' << s << endl;
}