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;
int n, k;
string s;
int f(string a, string b) {
  int t = 0;
  for (int i = 0; i < a.size(); i++)
    if (a[i] != b[i]) t++;
  return t;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k >> s;
  if (k == 2) {
    string a = "", b = "";
    string temp = "AB";
    while (a.size() < n) a += temp;
    temp = "BA";
    while (b.size() < n) b += temp;
    a.resize(n);
    b.resize(n);
    if (f(a, s) < f(b, s))
      cout << f(a, s) << "\n" << a << endl;
    else
      cout << f(b, s) << "\n" << b << endl;
    return 0;
  }
  int t = 0;
  s += '#';
  for (int i = 1; i < n; i++) {
    if (s[i - 1] != s[i]) continue;
    if (s[i] == s[i + 1] || s[i] == s[i - 1]) {
      if (s[i - 1] != 'A' && s[i + 1] != 'A')
        s[i] = 'A';
      else if (s[i - 1] != 'B' && s[i + 1] != 'B')
        s[i] = 'B';
      else
        s[i] = 'C';
      t++;
      i++;
    }
  }
  cout << t << endl;
  s.resize(n);
  cout << s << endl;
  return 0;
}