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() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  int cnt = 0;
  if (k > 2) {
    for (int i = 0; i < n; i++) {
      if (i < n - 1 && s[i + 1] == s[i]) {
        if (s[i + 2] == s[i]) {
          while (s[i + 1] == s[i]) {
            s[i + 1] = 'A' + rand() % k;
          }
        } else {
          while (s[i + 1] == s[i] || ((i > 0) && s[i] == s[i - 1])) {
            s[i] = 'A' + rand() % k;
          }
        }
        i++;
        cnt++;
      }
    }
  } else {
    int c1 = 0;
    string t = s;
    for (int i = 0; i < n; i++) {
      if (s[i] != ('A' + i % 2)) {
        c1++;
        t[i] = ('A' + i % 2);
      }
    }
    for (int i = 0; i < n; i++) {
      if (s[i] != ('A' + (i + 1) % 2)) {
        cnt++;
        s[i] = ('A' + (i + 1) % 2);
      }
    }
    if (c1 < cnt) {
      cnt = c1;
      s = t;
    }
  }
  cout << cnt << '\n' << s;
  return 0;
}