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 n, k;
string s;
void print() {
  string a = "", b = "";
  for (int i = 0; i < n; i++) {
    if (i & 1)
      a += 'A';
    else
      a += 'B';
  }
  for (int i = 0; i < n; i++) {
    if (i & 1)
      b += 'B';
    else
      b += 'A';
  }
  int one = 0, two = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] != a[i]) one++;
  }
  for (int i = 0; i < s.size(); i++) {
    if (s[i] != b[i]) two++;
  }
  if (one < two)
    cout << one << endl << a << endl;
  else
    cout << two << endl << b << endl;
}
int main() {
  while (cin >> n >> k) {
    int cnt = 0;
    cin >> s;
    if (k == 2) {
      print();
      return 0;
    }
    char en = 'A' + k - 1;
    for (int i = 0; i < s.size() - 1; i++) {
      if (s[i] == s[i + 1]) {
        cnt++;
        for (char ch = 'A'; ch <= en; ch++) {
          if (ch != s[i + 1] && ch != s[i + 2]) {
            s[i + 1] = ch;
            break;
          }
        }
      }
    }
    cout << cnt << endl;
    cout << s << endl;
  }
}