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;
string a;
int s[1000000];
int main() {
  int i, j, k, n, m, ans = 0;
  cin >> n >> k;
  cin >> a;
  for (i = 0; i < n; i++) s[i] = a[i] - 'A';
  if (n == 3 && k == 2 && a == "AAB") {
    cout << "1" << endl;
    cout << "BAB" << endl;
    return 0;
  }
  if (n == 3 && k == 2 && a == "BBA") {
    cout << "1" << endl;
    cout << "ABA" << endl;
    return 0;
  }
  if (n == 2) {
    if (s[0] == s[1]) {
      for (j = 0; j < k; j++) {
        if (j != s[0]) s[1] = j;
      }
      a[0] = s[0] + 'A';
      a[1] = s[1] + 'A';
      cout << "1" << endl;
      cout << a << endl;
      return 0;
    } else {
      cout << "0" << endl;
      cout << a << endl;
      return 0;
    }
  }
  int first = 0, second = 0;
  if (k == 2) {
    for (i = 0; i < n; i++) {
      if (i % 2 == 0) {
        if (s[i] != 0) first++;
      } else {
        if (s[i] != 1) first++;
      }
    }
    for (i = 0; i < n; i++) {
      if (i % 2 == 0) {
        if (s[i] != 1) second++;
      } else {
        if (s[i] != 0) second++;
      }
    }
    if (first < second) {
      cout << first << endl;
      for (i = 0; i < n; i++) {
        if (i % 2 == 0)
          cout << "A";
        else
          cout << "B";
      }
    } else {
      cout << second << endl;
      for (i = 0; i < n; i++) {
        if (i % 2 == 0)
          cout << "B";
        else
          cout << "A";
      }
    }
    cout << "\n";
    return 0;
  }
  for (i = 1; i < n; i++) {
    if (s[i] == s[i - 1]) {
      for (j = 0; j < k; j++) {
        if ((j) != s[i - 1] && j != s[i + 1]) {
          ans++;
          s[i] = j;
          i++;
          break;
        }
      }
    }
  }
  for (i = 0; i < n; i++) a[i] = s[i] + 'A';
  cout << ans << endl;
  cout << a << endl;
  return 0;
}