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 n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  if (k == 2) {
    int m1 = 0;
    int m2 = 0;
    for (int i = 0; i < n; i++) {
      if (i % 2 && s[i] == 'A')
        m1++;
      else if (!(i % 2) && s[i] == 'B')
        m1++;
      if (i % 2 && s[i] == 'B')
        m2++;
      else if (!(i % 2) && s[i] == 'A')
        m2++;
    }
    if (m1 <= m2) {
      for (int i = 0; i < n; i++) s[i] = (i % 2 ? 'B' : 'A');
    } else {
      for (int i = 0; i < n; i++) s[i] = (i % 2 ? 'A' : 'B');
    }
    cout << min(m1, m2) << endl;
    cout << s << endl;
    return 0;
  }
  int anw = 0;
  s += '0';
  for (int i = 1; i < n; i++) {
    if (s[i - 1] == s[i]) {
      anw++;
      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';
    }
  }
  s = s.substr(0, n);
  cout << anw << endl << s << endl;
}