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;
const long long mod = 1e9 + 7;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  while (t--) {
    long long n, k, cnt = 0, cnt1 = 0, cnt2 = 0;
    cin >> n >> k;
    string s, s1 = "", s2 = "";
    cin >> s;
    if (k == 2) {
      for (int i = 0; i < n; i++) {
        if (i % 2) {
          s1 += "A";
          s2 += "B";
        } else {
          s1 += "B";
          s2 += "A";
        }
      }
      for (int i = 0; i < n; i++) {
        if (s[i] != s1[i]) cnt1++;
        if (s[i] != s2[i]) cnt2++;
      }
      if (cnt1 < cnt2) {
        cout << cnt1 << endl;
        cout << s1 << endl;
      } else {
        cout << cnt2 << endl;
        cout << s2 << endl;
      }
      continue;
    }
    for (int i = 1; i < n; i++) {
      if (s[i - 1] == s[i]) {
        for (int j = 0; j < k; j++) {
          char c = 'A' + j;
          if (c != s[i - 1] && c != s[i + 1]) {
            s[i] = c;
            cnt++;
            break;
          }
        }
      }
    }
    cout << cnt << endl;
    cout << s << endl;
  }
}