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;
void solution() {
  int n, k, i, j, sum = 0;
  cin >> n >> k;
  string s;
  cin >> s;
  if (k == 2) {
    int c1 = 0, c2 = 0;
    for (i = 0; i < n; i++) {
      c1 += abs(s[i] - 65 - (i % 2));
      c2 += abs(s[i] - 65 - (i + 1) % 2);
    }
    if (c1 < c2) {
      cout << c1 << "\n";
      for (i = 0; i < n; i++) cout << (char)(65 + i % 2);
    } else {
      cout << c2 << "\n";
      for (i = 0; i < n; i++) cout << (char)(65 + (i + 1) % 2);
    }
    return;
  }
  for (i = 0; i < n - 1; i++) {
    char c = 'A';
    if (s[i] == s[i + 1]) {
      sum++;
      while (s[i + 2] == c || s[i] == c) {
        c = (char)(c + 1);
      }
      s[i + 1] = c;
    }
  }
  cout << sum << "\n" << s;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solution();
  }
  cerr << "\n" << (float)clock() / CLOCKS_PER_SEC * 1000 << " ms" << endl;
  return 0;
}