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() {
  char c[600000];
  int str, k;
  cin >> str >> k >> c;
  int i, min = 0;
  if (k == 2) {
    int x = 0;
    for (i = 0; i < str; ++i)
      if (i % 2 == 0) {
        if (c[i] == 'B')
          ++x;
        else
          ++min;
      } else if (c[i] == 'A')
        ++x;
      else
        ++min;
    if (x < min) {
      cout << x << endl;
      for (i = 0; i < str; ++i)
        if (i % 2 == 0)
          cout << 'A';
        else
          cout << 'B';
    } else {
      cout << min << endl;
      for (i = 0; i < str; ++i)
        if (i % 2 != 0)
          cout << 'A';
        else
          cout << 'B';
    }
    return 0;
  }
  if (strlen(c) > 2) {
    for (i = 0; c[i + 2]; ++i) {
      if (c[i] == c[i + 1] && c[i] == c[i + 2]) {
        c[i + 1] = 'A', ++min;
      }
      if (c[i] == c[i + 1] && c[i] == c[i + 2]) c[i + 1] = 'B';
    }
  }
  int we;
  for (i = 0; c[i + 1]; ++i) {
    we = 0;
    if (c[i] == c[i + 1]) {
      we = 1;
      c[i + 1] = 'A', ++min;
    }
    if (we && (c[i] == c[i + 1] || c[i + 1] == c[i + 2])) c[i + 1] = 'B';
    if (we && (c[i] == c[i + 1] || c[i + 1] == c[i + 2])) c[i + 1] = 'C';
  }
  cout << min << endl;
  cout << c;
  return 0;
}