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;
char a[500003], d[5000003], e[5000003];
int n;
queue<int> q;
int convert() {
  int c = 0, i;
  for (i = 1; i <= n; i++) {
    if (a[i] != d[i]) c++;
  }
  return c;
}
int main() {
  int i, j, k, x, y, z, f = 1;
  char c;
  cin >> n >> k;
  scanf("%s", a);
  for (i = n; i >= 1; i--) {
    a[i] = a[i - 1];
  }
  if (k == 2) {
    for (i = 1; i <= n; i++) {
      if (i % 2 == 1) {
        d[i] = 'A';
        e[i] = 'A';
      } else {
        d[i] = 'B';
        e[i] = 'B';
      }
    }
    x = convert();
    for (i = 1; i <= n; i++) {
      if (i % 2 == 0)
        d[i] = 'A';
      else
        d[i] = 'B';
    }
    y = convert();
    if (x < y) {
      cout << x << endl;
      for (i = 1; i <= n; i++) printf("%c", e[i]);
    } else {
      cout << y << endl;
      for (i = 1; i <= n; i++) printf("%c", d[i]);
    }
    return 0;
  }
  a[n + 1] = '\0';
  for (i = 1; i <= n; i++) {
    if (f == 1) {
      if (a[i] == a[i + 1]) {
        q.push(i);
        f = 2;
      }
    }
    if (f == 2) {
      if (a[i] != a[i + 1]) {
        q.push(i);
        f = 1;
      }
    }
  }
  int count = 0;
  while (!q.empty()) {
    x = q.front();
    q.pop();
    y = q.front();
    q.pop();
    c = a[x];
    if (c == 'A') {
      if (a[y + 1] == 'B')
        c = 'C';
      else
        c = 'B';
    } else if (c == 'B') {
      if (a[y + 1] == 'A')
        c = 'C';
      else
        c = 'A';
    } else {
      if (a[y + 1] == 'A')
        c = 'B';
      else
        c = 'A';
    }
    for (i = x + 1; i <= y; i = i + 2) {
      a[i] = c;
      count++;
    }
  }
  cout << count << endl;
  for (i = 1; i <= n; i++) printf("%c", a[i]);
}