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;
bool debug_cerr = 1;
const long long mod = 1000000007;
template <typename T>
void printarr(T arr[], int n) {
  if (!debug_cerr) {
  } else
    cerr << "Array ";
  for (int i = 0; i < n; i++) {
    if (!debug_cerr) {
    } else
      cerr << arr[i] << " ";
  }
  if (!debug_cerr) {
  } else
    cerr << endl;
}
template <typename T>
void printarr(T arr) {
  if (!debug_cerr) {
  } else
    cerr << "AnyList ";
  for (auto it = arr.begin(); it != arr.end(); it++) {
    if (!debug_cerr) {
    } else
      cerr << *it << " ";
  }
  if (!debug_cerr) {
  } else
    cerr << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long test, ti, i, j, n, m, k;
  string s;
  cin >> n >> k;
  cin >> s;
  s = '1' + s + '1';
  int cnt = 0;
  if (k == 2) {
    int codd = 0, ceven = 0;
    bool odd = 0;
    string sodd = s, seven = s;
    for (i = (1); i <= (n); i++) {
      char c = 'A' + (i % 2 == 0);
      if (s[i] != c) {
        codd++;
        sodd[i] = c;
      }
    }
    for (i = (1); i <= (n); i++) {
      char c = 'A' + (i % 2 == 1);
      if (s[i] != c) {
        ceven++;
        seven[i] = c;
      }
    }
    if (codd < ceven)
      odd = 1, cnt = codd, s = sodd;
    else
      odd = 0, cnt = ceven, s = seven;
    goto fin;
  }
  for (i = (1); i <= (n); i++) {
    if (s[i - 1] == s[i]) {
      cnt++;
      while (s[i] == s[i - 1] || s[i] == s[i + 1]) {
        s[i] = (s[i] - 'A' + 1) % k + 'A';
      }
    }
  }
fin:
  cout << cnt << "\n";
  for (i = (1); i <= (n); i++) {
    cout << s[i];
  }
  cout << "\n";
  return 0;
}