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;
  char a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  if (k != 2) {
    int num = 0;
    for (int i = 1; i < n; i++)
      if (a[i] == a[i - 1]) {
        num++;
        if (a[i - 1] != 'A' && (i + 1 >= n || a[i + 1] != 'A')) a[i] = 'A';
        if (a[i - 1] != 'B' && (i + 1 >= n || a[i + 1] != 'B')) a[i] = 'B';
        if (a[i - 1] != 'C' && (i + 1 >= n || a[i + 1] != 'C')) a[i] = 'C';
      }
    cout << num << "\n";
    for (int i = 0; i < n; i++) cout << a[i];
    return 0;
  }
  int x = 0, y = 0;
  for (int i = 0; i < n; i++)
    if (i % 2) {
      if (a[i] == 'A')
        x++;
      else
        y++;
    } else {
      if (a[i] == 'A')
        y++;
      else
        x++;
    }
  char A, B;
  if (x > y) {
    A = 'A';
    B = 'B';
    cout << y << "\n";
  } else {
    A = 'B';
    B = 'A';
    cout << x << "\n";
  }
  for (int i = 0; i < n; i++) cout << (i % 2 ? A : B);
  return 0;
}