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;
char col(long long int i, long long int j, long long int l) {
  for (long long int k = 0; k < 3; k++) {
    if (i != k && j != k) {
      return 'A' + k;
    }
  }
  return '0';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  if (k == 2) {
    string a = "", b = "";
    long long int la = 0, lb = 1;
    la = 0, lb = 0;
    for (long long int i = 0; i < s.length(); i++) {
      if (s[i] == 'A') {
        if (i % 2 == 1) {
          la++;
        } else {
          lb++;
        }
      } else {
        if (i % 2 == 0) {
          la++;
        } else {
          lb++;
        }
      }
    }
    if (lb < la) {
      cout << lb << endl;
      for (long long int i = 0; i < s.length(); i++) {
        if (i % 2 == 0) {
          cout << 'B';
        } else {
          cout << 'A';
        }
      }
    } else {
      cout << la << endl;
      for (long long int i = 0; i < s.length(); i++) {
        if (i % 2 == 1) {
          cout << 'B';
        } else {
          cout << 'A';
        }
      }
    }
  } else {
    long long int cn = 0;
    for (long long int i = 1; i < s.length(); i++) {
      if (s[i] == s[i - 1]) {
        if (i == s.length() - 1) {
          s[i] = col(s[i - 1] - 'A', s[i - 1] - 'A', k);
          cn++;
          break;
        }
        s[i] = col(s[i - 1] - 'A', s[i + 1] - 'A', k);
        cn++;
      }
    }
    cout << cn << endl;
    cout << s;
  }
  return 0;
}