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;
  string s;
  cin >> s;
  for (int i = 0; i < k; i++) {
  }
  if (k == 2) {
    int a = 0, b = 0;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) {
        if (s[i] == 'B') {
          a++;
        } else {
          b++;
        }
      } else {
        if (s[i] == 'A') {
          a++;
        } else {
          b++;
        }
      }
    }
    int x = min(a, b);
    cout << x << '\n';
    if (x == a) {
      for (int i = 0; i < n; i++) {
        if (i % 2 == 0) {
          cout << 'A';
        } else {
          cout << 'B';
        }
      }
    } else {
      for (int i = 0; i < n; i++) {
        if (i % 2 == 0) {
          cout << 'B';
        } else {
          cout << 'A';
        }
      }
    }
  } else {
    int i = 0, ans = 0, j;
    while (i < n) {
      for (j = i; j < n; j++) {
        if (s[j] != s[i]) {
          break;
        }
      }
      ans = ans + (j - i) / 2;
      if ((j - i) % 2 == 1) {
        for (int k = i + 1; k < j; k = k + 2) {
          if (s[i] == 'A') {
            s[k] = 'B';
          } else {
            s[k] = 'A';
          }
        }
      } else {
        char ch;
        if (s[i] != 'A' && s[j] != 'A') {
          ch = 'A';
        } else if (s[i] != 'B' && s[j] != 'B') {
          ch = 'B';
        } else {
          ch = 'C';
        }
        for (int k = i + 1; k < j; k = k + 2) {
          s[k] = ch;
        }
      }
      i = j;
    }
    cout << ans << '\n';
    cout << s;
  }
  return 0;
}