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 s[1234567];
int main() {
  ios ::sync_with_stdio(false);
  int n, k;
  int ans = 0;
  cin >> n >> k;
  cin >> s;
  if (k == 2) {
    int x = 0, y = 0, x1 = 0, y1 = 0;
    for (int i = 0; i < n; ++i) {
      if (i % 2 == 0 && s[i] != 'A') x++;
      if (i % 2 == 1 && s[i] != 'B') y++;
    }
    for (int i = 0; i < n; ++i) {
      if (i % 2 == 0 && s[i] != 'B') x1++;
      if (i % 2 == 1 && s[i] != 'A') y1++;
    }
    int m = min(x + y, x1 + y1);
    ans = m;
    if (m == x + y) {
      for (int i = 0; i < n; i += 2) {
        s[i] = 'A';
        s[i + 1] = 'B';
      }
    } else {
      for (int i = 0; i < n; i += 2) {
        s[i] = 'B';
        s[i + 1] = 'A';
      }
    }
  } else {
    for (int i = 0; i + 1 < n; ++i) {
      if (s[i] == s[i + 1]) {
        char w = 'A';
        if (s[i] == w || s[i + 2] == w) w++;
        if (s[i] == w || s[i + 2] == w) w++;
        ans++;
        s[i + 1] = w;
      }
    }
  }
  cout << ans << endl;
  for (int i = 0; i < n; i++) cout << s[i];
  cout << endl;
  return 0;
}