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() {
  clock_t begin = clock();
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  if (k == 2) {
    string temp = "";
    for (int i = 0; i < n; i++) {
      if (i & 1)
        temp.push_back('B');
      else
        temp.push_back('A');
    }
    int ct = 0;
    for (int i = 0; i < n; i++) {
      if (temp[i] != s[i]) ct++;
    }
    int tt = n - ct;
    if (ct < tt) {
      cout << ct << endl;
      cout << temp << endl;
    } else {
      cout << tt << endl;
      for (int i = 0; i < n; i++) {
        if (i & 1)
          cout << "A";
        else
          cout << "B";
      }
    }
    return 0;
  } else {
    int ans = 0;
    s.push_back('Z');
    for (int i = 1; i < n; i++) {
      if (s[i] == s[i - 1]) {
        ans++;
        for (int j = 0; j < k; j++) {
          if (s[i - 1] - 'A' != j && s[i + 1] - 'A' != j) {
            s[i] = (char)'A' + j;
            break;
          }
        }
      }
    }
    cout << ans << endl;
    s.pop_back();
    cout << s << endl;
  }
  return 0;
}