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 argc, char const *argv[]) {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  char clr[k];
  for (int i = 0; i < k; i++) {
    clr[i] = 65 + i;
  }
  if (k > 2) {
    int noc = 0;
    for (int i = 1; i < s.length() - 1; i++) {
      if (s[i] == s[i - 1]) {
        for (int j = 0; j < k; j++) {
          if (clr[j] != s[i + 1] && clr[j] != s[i - 1]) {
            s[i] = clr[j];
            noc++;
            break;
          }
        }
      }
    }
    n = s.length();
    if (s[0] == s[1]) {
      int i = 0;
      for (int j = 0; j < k; j++) {
        if (clr[j] != s[i + 1]) {
          s[i] = clr[j];
          noc++;
          break;
        }
      }
    }
    if (s[n - 1] == s[n - 2]) {
      for (int j = 0; j < k; j++) {
        if (clr[j] != s[n - 2]) {
          s[n - 1] = clr[j];
          noc++;
          break;
        }
      }
    }
    cout << noc << endl;
    cout << s;
  } else {
    string s1, s2;
    s1 = s;
    s2 = s;
    s1[0] = 'A';
    s2[0] = 'B';
    for (int i = 1; i < s.length(); i++) {
      if (s1[i - 1] == 'A')
        s1[i] = 'B';
      else
        s1[i] = 'A';
      if (s2[i - 1] == 'A')
        s2[i] = 'B';
      else
        s2[i] = 'A';
    }
    int n1 = 0;
    int n2 = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] != s1[i]) n1++;
      if (s[i] != s2[i]) n2++;
    }
    if (n1 < n2) {
      cout << n1 << endl << s1;
    } else {
      cout << n2 << endl << s2;
    }
  }
  return 0;
}