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;
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k, c = 0;
  cin >> n >> k;
  string s;
  cin >> s;
  if (n == 1) {
    cout << 0 << "\n" << s;
    exit(0);
  }
  if (k == 2) {
    int d = 0;
    for (long long i = 0; i < n; i++) {
      if (s[i] == 'A' && i % 2 == 0) {
        c++;
      }
      if (s[i] == 'B' && i % 2 != 0) {
        c++;
      }
    }
    for (long long i = 0; i < n; i++) {
      if (s[i] == 'A' && i % 2 != 0) {
        d++;
      }
      if (s[i] == 'B' && i % 2 == 0) {
        d++;
      }
    }
    if (c < d) {
      cout << c << "\n";
      for (long long i = 0; i < n; i++) {
        if (i % 2 == 0) {
          cout << 'B';
        } else {
          cout << 'A';
        }
      }
    } else {
      cout << d << "\n";
      for (long long i = 0; i < n; i++) {
        if (i % 2 == 0) {
          cout << 'A';
        } else {
          cout << 'B';
        }
      }
    }
  } else {
    for (long long i = 1; i < n; i++) {
      if (s[i] == s[i - 1]) {
        c++;
        if (i == n - 1) {
          char ans;
          for (int j = 0; j < k; j++) {
            char ch = char(65 + j);
            if (ch != s[i - 1]) {
              ans = ch;
              break;
            }
          }
          s[i] = ans;
        } else {
          char ans;
          for (int j = 0; j < k; j++) {
            char ch = char(65 + j);
            if (ch == s[i - 1] || ch == s[i + 1]) {
              continue;
            } else {
              ans = ch;
              break;
            }
          }
          s[i] = ans;
        }
      }
    }
    cout << c << "\n" << s << "\n";
  }
}