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;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
int n, k;
string str;
int ret = 0;
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> n >> k >> str;
  int len = str.size();
  if (k == 2) {
    int tmp[2] = {0, 0};
    string tmpstr = "";
    for (int i = 0; i < len; i++) {
      if (i % 2 == 0) {
        if (str[i] == 'A') {
          tmp[0]++;
        } else {
          tmp[1]++;
        }
      } else {
        if (str[i] == 'A') {
          tmp[1]++;
        } else {
          tmp[0]++;
        }
      }
    }
    if (tmp[0] < tmp[1]) {
      ret = tmp[0];
      for (int i = 0; i < len; i++) {
        tmpstr += (i % 2) ? 'A' : 'B';
      }
    } else {
      ret = tmp[1];
      for (int i = 0; i < len; i++) {
        tmpstr += (i % 2) ? 'B' : 'A';
      }
    }
    str = tmpstr;
  } else {
    for (int i = 0; i < len; i++) {
      int p = i;
      if (str[p] == str[p + 1]) {
        ret++;
        for (int i = 0; i < k; i++) {
          if (p + 2 < len) {
            if (str[p] != i + 'A' && str[p + 2] != i + 'A') {
              str[p + 1] = i + 'A';
            }
          } else {
            if (str[p] != i + 'A') {
              str[p + 1] = i + 'A';
            }
          }
        }
      }
    }
  }
  cout << ret << endl;
  cout << str << endl;
}