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() {
  long long n, k;
  cin >> n >> k;
  string s, s1;
  cin >> s;
  if (k == 2) {
    long long cnt1 = 0, cnt2 = 0;
    for (long long i = 0; i < n; i++) {
      if (i % 2 == 0 && s[i] == 'A') cnt1++;
      if (i % 2 == 1 && s[i] == 'B') cnt1++;
    }
    for (long long i = 0; i < n; i++) {
      if (i % 2 == 0 && s[i] == 'B') cnt2++;
      if (i % 2 == 1 && s[i] == 'A') cnt2++;
    }
    cout << min(cnt1, cnt2) << endl;
    if (cnt1 < cnt2) {
      for (long long i = 0; i < n; i++) {
        if (i % 2 == 0)
          cout << 'B';
        else
          cout << 'A';
      }
    } else {
      for (long long i = 0; i < n; i++) {
        if (i % 2 == 1)
          cout << 'B';
        else
          cout << 'A';
      }
    }
  } else {
    long long res = 0, cnt = 1;
    for (long long i = 1; i < n; i++) {
      if (s[i] == s[i - 1])
        cnt++;
      else
        res += (cnt / 2), cnt = 1;
    }
    res += cnt / 2;
    cout << res << endl;
    for (long long i = 1; i < n; i++) {
      if (s[i] == s[i - 1] && s[i] == 'A') {
        if (i < n - 1 && s[i + 1] == 'B')
          s[i] = 'C';
        else
          s[i] = 'B';
      } else if (s[i] == s[i - 1] && s[i] == 'B') {
        if (i < n - 1 && s[i + 1] == 'C')
          s[i] = 'A';
        else
          s[i] = 'C';
      } else if (s[i] == s[i - 1]) {
        if (i < n - 1 && s[i + 1] == 'A')
          s[i] = 'B';
        else
          s[i] = 'A';
      }
    }
    cout << s;
  }
  cout << endl;
  return 0;
}