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 int MAXN = 5e5 + 5;
int A[MAXN], B[MAXN], C[MAXN];
int main() {
  ios::sync_with_stdio(false);
  char ch;
  int n, k, ans = 0;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> ch;
    A[i] = ch - 'A' + 1;
  }
  A[0] = 0;
  A[n + 1] = 0;
  if (k > 2) {
    for (int i = 2; i <= n; i++) {
      if (A[i] == A[i - 1]) {
        ans++;
        for (int j = 1; j <= k; j++) {
          if (j != A[i - 1] && j != A[i + 1]) {
            A[i] = j;
            break;
          }
        }
      }
    }
    cout << ans << endl;
    for (int i = 1; i <= n; i++) cout << (char)(A[i] + 'A' - 1);
    cout << endl;
  } else {
    int ans1 = 0, ans2 = 0, a;
    for (int i = 1; i <= n; i++) {
      a = (i % 2) + 1;
      if (A[i] != a)
        ans1++;
      else
        ans2++;
    }
    if (ans1 < ans2) {
      cout << ans1 << endl;
      for (int i = 1; i <= n; i++) cout << (char)(((i % 2) + 1) + 'A' - 1);
    } else {
      cout << ans2 << endl;
      for (int i = 1; i <= n; i++) cout << (char)(((i + 1) % 2 + 1) + 'A' - 1);
    }
  }
  return 0;
}