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;
void k2() {
  int dema = 0, demb = 0;
  for (int i = 0; i < n; i++) {
    if (i % 2 == 0) {
      if (S[i] == 'A')
        demb++;
      else
        dema++;
    } else {
      if (S[i] == 'B')
        demb++;
      else
        dema++;
    }
  }
  if (dema > demb) {
    cout << demb << endl;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0)
        cout << "B";
      else
        cout << "A";
    }
  } else {
    cout << dema << endl;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0)
        cout << "A";
      else
        cout << "B";
    }
  }
}
void k1() {
  int res = 0, cnt = 0;
  ;
  for (int i = 1; i < n; i++) {
    if (S[i] == S[i - 1]) {
      cnt++;
      if (i + 1 < n) {
        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';
      } else {
        if (S[i - 1] != 'A') {
          S[i] = 'A';
        } else if (S[i - 1] != 'B') {
          S[i] = 'B';
        } else
          S[i] = 'C';
      }
    }
  }
  cout << cnt << endl << S;
}
int main() {
  cin >> n >> k;
  cin >> S;
  if (k == 2)
    k2();
  else
    k1();
  return 0;
}