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 ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  if (k == 2) {
    int a = 0, b = 0, c1 = 0, c2 = 0;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0 && s[i] != 'A')
        a++;
      else if (i % 2 && s[i] != 'B')
        b++;
    }
    c1 = a + b;
    a = 0;
    b = 0;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0 && s[i] != 'B')
        b++;
      else if (i % 2 && s[i] != 'A')
        a++;
    }
    c2 = a + b;
    if (c1 < c2) {
      cout << c1 << endl;
      for (int i = 0; i < n; i++) {
        if (i % 2 == 0)
          cout << 'A';
        else
          cout << 'B';
      }
    } else {
      cout << c2 << endl;
      for (int i = 0; i < n; i++) {
        if (i % 2 == 0)
          cout << 'B';
        else
          cout << 'A';
      }
    }
  } else {
    s += '.';
    for (int i = 1; i < n; i++) {
      if (s[i] == s[i - 1]) {
        for (int c = 0; c < k; c++) {
          char x = 'A' + c;
          if (s[i - 1] != x && x != s[i + 1]) {
            s[i] = x;
            ans++;
            break;
          }
        }
      }
    }
    cout << ans << "\n";
    s.pop_back();
    cout << s;
  }
}