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;
int a[500005];
int memo[500005][30];
int solve(int i, int p) {
  if (i == n) {
    return 0;
  }
  if (memo[i][p] != -1) {
    return memo[i][p];
  }
  int ret = 1e9;
  if (a[i] != p) {
    ret = min(ret, solve(i + 1, a[i]));
  }
  for (int q = 1; q <= min(7, k); q++) {
    if (q != p) {
      if (q == a[i]) {
        ret = min(ret, solve(i + 1, q));
      } else {
        ret = min(ret, 1 + solve(i + 1, q));
      }
    }
  }
  return memo[i][p] = ret;
}
void bulid(int i, int p) {
  if (i == n) {
    return;
  }
  int ret = 1e9;
  int c = 0;
  if (a[i] != p) {
    ret = min(ret, solve(i + 1, a[i]));
    c = a[i];
  }
  for (int q = 1; q <= min(7, k); q++) {
    if (q != p) {
      if (q == a[i]) {
        int y = solve(i + 1, q);
        if (ret > y) {
          ret = y;
          c = q;
        }
      } else {
        int y = 1 + solve(i + 1, q);
        if (ret > y) {
          ret = y;
          c = q;
        }
      }
    }
  }
  cout << (char)(c + 'A' - 1);
  return bulid(i + 1, c);
}
int main() {
  memset(memo, -1, sizeof memo);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    char x;
    cin >> x;
    int d = x - 'A' + 1;
    a[i] = d;
  }
  cout << solve(0, 0) << endl;
  bulid(0, 0);
  return 0;
}