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;
string s;
char f(char c) {
  if (c == 'A') return 'B';
  return 'A';
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  cin >> s;
  int ctr = 0;
  if (k == 2) {
    string ss = s;
    ss[0] = f(s[0]);
    int a = 1, b = 0;
    for (int i = 1; i < s.size(); i++) {
      if (ss[i] == ss[i - 1]) {
        a++;
        ss[i] = f(ss[i]);
      }
      if (s[i] == s[i - 1]) {
        b++;
        s[i] = f(s[i]);
      }
    }
    if (a < b)
      cout << a << endl << ss << endl;
    else
      cout << b << endl << s << endl;
  } else {
    int ctr = 0;
    for (int i = 0; i < s.size(); i++) {
      if (i + 1 < s.size() && s[i] == s[i + 1]) {
        ctr++;
        s[i + 1] = 'A';
        if (i + 2 < s.size()) {
          while (s[i + 1] == s[i] || s[i + 1] == s[i + 2]) s[i + 1]++;
        } else {
          while (s[i + 1] == s[i]) s[i + 1]++;
        }
        i++;
      } else if (i > 0 && s[i] == s[i - 1]) {
        ctr++;
        s[i] = 'A';
        while (s[i] == s[i - 1]) s[i]++;
      }
    }
    cout << ctr << endl << s << endl;
  }
}