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;
bool b[5 * 100000];
void doit() {
  int res = 0;
  char pat = s[0];
  int ct = 0;
  for (int i = 0; i < s.length(); ++i) {
    if (s[i] == pat)
      ct++;
    else {
      if (ct > 1) {
        if (ct % 2 == 0) {
          for (int pos = i - 1, cant = ct / 2; cant > 0; cant--) {
            b[pos] = 1;
            pos -= 2;
          }
        } else {
          for (int pos = i - 2, cant = ct / 2; cant > 0; cant--) {
            b[pos] = 1;
            pos -= 2;
          }
        }
        res += ct / 2;
      }
      ct = 1;
      pat = s[i];
    }
  }
  if (ct > 1) {
    if (ct % 2 == 0) {
      for (int pos = s.length() - 1, cant = ct / 2; cant > 0; cant--) {
        b[pos] = 1;
        pos -= 2;
      }
    } else {
      for (int pos = s.length() - 2, cant = ct / 2; cant > 0; cant--) {
        b[pos] = 1;
        pos -= 2;
      }
    }
    res += ct / 2;
  }
  for (int i = 0; i < s.length(); ++i) {
    if (b[i]) {
      char c = 'A';
      if (i == 0) {
        while (c == s[1]) c++;
        s[i] = c;
      } else {
        if (i == (s.length() - 1)) {
          while (c == s[i - 1]) c++;
          s[i] = c;
        } else {
          while (c == s[i - 1] || c == s[i + 1]) c++;
          s[i] = c;
        }
      }
    }
  }
  cout << res << endl;
  cout << s << endl;
}
int main() {
  cin >> n >> k;
  cin >> s;
  if (k > 2)
    doit();
  else {
    string s1 = "", s2 = "";
    int ct1 = 0, ct2 = 0;
    for (int i = 0; i < s.length(); ++i) {
      s1 += (i % 2) + 'A';
      if (s[i] != s1[i]) ct1++;
      s2 += ((i % 2) + 1) % 2 + 'A';
      if (s[i] != s2[i]) ct2++;
    }
    if (ct1 < ct2) {
      cout << ct1 << endl;
      cout << s1;
    } else {
      cout << ct2 << endl;
      cout << s2 << endl;
    }
  }
}