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 int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  map<char, long long int> check;
  map<long long int, char> make_pair;
  string t = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
  for (long long int i = 0; i < t.size(); i++) {
    check[t[i]] = i + 1;
    make_pair[i + 1] = t[i];
  }
  if (k == 2) {
    long long int a = 0, b = 0;
    for (long long int i = 0; i < n; i++) {
      if (s[i] - 'A' != i % 2)
        a++;
      else
        b++;
    }
    cout << min(a, b) << endl;
    if (a <= b) {
      for (long long int i = 0; i < n; i++) {
        if (i % 2)
          cout << "B";
        else
          cout << "A";
      }
    } else {
      for (long long int i = 0; i < n; i++) {
        if (i % 2)
          cout << "A";
        else
          cout << "B";
      }
    }
    return 0;
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    if (s[i] == s[i + 1]) {
      long long int x = min(check[s[i + 1]], check[s[i + 2]]) - 1;
      long long int y = min(check[s[i + 1]], check[s[i + 2]]) + 1;
      long long int z = max(check[s[i + 1]], check[s[i + 2]]) + 1;
      if (x >= 1 && x <= k) {
        s[i + 1] = make_pair[x];
      } else if (make_pair[y] != s[i + 1] && make_pair[y] != s[i + 2] && y <= k)
        s[i + 1] = make_pair[y];
      else if (z <= k)
        s[i + 1] = make_pair[z];
      else {
        s[i] = make_pair[y];
      }
      ans++;
    }
  }
  cout << ans << endl << s << endl;
}