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 INF = (1ll << 62) - 1;
const int inf = (1 << 30) - 1;
const int M = 500007;
int cnt;
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  string st;
  cin >> n >> k;
  cin >> st;
  if (k == 2) {
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) {
        if (st[i] != 'A') cnt++;
      } else {
        if (st[i] != 'B') cnt++;
      }
    }
    if (cnt <= n - cnt) {
      cout << cnt << endl;
      for (int i = 0; i < n; i++) {
        if (i % 2 == 0)
          cout << "A";
        else
          cout << "B";
      }
    } else {
      cout << n - cnt << endl;
      for (int i = 0; i < n; i++) {
        if (i % 2 == 0)
          cout << "B";
        else
          cout << "A";
      }
    }
  } else {
    for (int i = 1; i < n; i++) {
      if (st[i - 1] == st[i]) {
        for (char ch = 'A';; ch++) {
          if (st[i] != ch && st[i + 1] != ch) {
            st[i] = ch;
            cnt++;
            break;
          }
        }
      }
    }
    cout << cnt << endl;
    cout << st << endl;
    return 0;
  }
}