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;
using namespace std;
const long long MOD = 1e9 + 7;
const int INF = 1e9;
int n, k;
string s;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  cin >> s;
  if (k == 2) {
    string temp1 = "";
    string temp2 = "";
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) {
        temp1 += 'B';
        temp2 += 'A';
      } else {
        temp1 += 'A';
        temp2 += 'B';
      }
    }
    int diff1 = 0, diff2 = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] != temp1[i]) diff1++;
    }
    for (int i = 0; i < n; i++) {
      if (s[i] != temp2[i]) diff2++;
    }
    s = diff1 > diff2 ? temp2 : temp1;
    cout << min(diff1, diff2) << "\n";
    cout << s << "\n";
    return 0;
  }
  int ans = 0;
  for (int i = 1; i <= n - 1; i++) {
    if (s[i] == s[i - 1]) {
      ans++;
      s[i] = 'A';
      while (s[i] == s[i - 1] or s[i] == s[i + 1]) s[i]++;
    }
  }
  cout << ans << "\n";
  cout << s << "\n";
  return 0;
}