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 a[100];
int main() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  if (k == 2) {
    int s1, s2;
    s1 = s2 = 0;
    for (int i = 0; i < n; i++)
      if (s[i] == 'A' ^ (i % 2 == 0))
        s1++;
      else
        s2++;
    for (int i = 0; i < n; i++)
      if (s1 > s2 ^ (i % 2 == 0))
        s[i] = 'A';
      else
        s[i] = 'B';
    cout << min(s1, s2) << endl << s << endl;
    return 0;
  }
  int res = 0;
  for (int i = 1; i < n; i++)
    if (s[i] == s[i - 1]) {
      res++;
      s[i] = '0';
    }
  for (int i = 1; i < n - 1; i++)
    if (s[i] == '0') {
      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';
    }
  if (s[n - 1] == '0' && s[n - 2] == 'A') s[n - 1] = 'B';
  if (s[n - 1] == '0') s[n - 1] = 'A';
  cout << res << endl << s << endl;
  return 0;
}