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 n, p;
string s;
inline int gn(int x) { return s[x] - int('A'); }
int go(int a, int b) {
  for (int i = 0; i < p; i++) {
    if (i != a && i != b) return i;
  }
}
int main() {
  cin >> n >> p >> s;
  if (p == 2) {
    int ans1 = 0;
    int ans2 = 0;
    for (int i = 0; i < s.length(); i++) {
      if ((i & 1) && s[i] == 'B')
        ans2++;
      else if ((i & 1) && s[i] == 'A')
        ans1++;
      else if (!(i & 1) && s[i] == 'A')
        ans2++;
      else
        ans1++;
    }
    cout << min(ans1, ans2) << endl;
    if (ans1 < ans2) {
      for (int i = 0; i < s.length(); i++) {
        if (i & 1)
          cout << 'B';
        else
          cout << 'A';
      }
    } else {
      for (int i = 0; i < s.length(); i++) {
        if (i & 1)
          cout << 'A';
        else
          cout << 'B';
      }
    }
    return 0;
  }
  int ans = 0;
  for (int i = 0; i < s.length(); i++) {
    int j = i;
    while (j + 1 < s.length() && s[j + 1] == s[j]) j++;
    if (j == i) continue;
    if ((j - i + 1) & 1) {
      int nc = go(-1, gn(i));
      for (int k = i + 1; k <= j; k += 2) {
        s[k] = char(nc + int('A'));
        ans++;
      }
    } else {
      int nc;
      if (j == s.length() - 1)
        nc = go(-1, gn(i));
      else
        nc = go(gn(j + 1), gn(i));
      for (int k = i + 1; k <= j; k += 2) {
        s[k] = char(nc + int('A'));
        ans++;
      }
    }
    i = j;
  }
  cout << ans << endl;
  cout << s;
  return 0;
}