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 s[500005], ans[500005];
int k, n;
int dp[500005][2];
int main() {
  scanf("%d%d%s", &n, &k, s);
  if (k == 2) {
    int tol1 = 0, tol2 = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] != i % 2 + 'A') tol1++;
      if (s[i] != (i + 1) % 2 + 'A') tol2++;
    }
    printf("%d\n", min(tol1, tol2));
    int g = tol1 < tol2 ? 0 : 1;
    for (int i = 0; i < n; i++) printf("%c", 'A' + (i + g) % 2);
    printf("\n");
    return 0;
  }
  dp[0][0] = 0;
  dp[0][1] = 1;
  for (int i = 1; s[i]; i++) {
    dp[i][1] = min(dp[i - 1][0], dp[i - 1][1]) + 1;
    if (s[i] == s[i - 1])
      dp[i][0] = dp[i - 1][1];
    else
      dp[i][0] = min(dp[i - 1][0], dp[i - 1][1]);
  }
  printf("%d\n", min(dp[n - 1][0], dp[n - 1][1]));
  int j = dp[n - 1][0] < dp[n - 1][1] ? 0 : 1;
  for (int i = n - 1; i >= 0; i--) {
    if (j == 1) {
      ans[i] = '?';
      if (i == 0) break;
      j = dp[i - 1][0] < dp[i - 1][1] ? 0 : 1;
    } else {
      ans[i] = s[i];
      if (i == 0) break;
      if (s[i] == s[i - 1])
        j = 1;
      else
        j = dp[i - 1][0] < dp[i - 1][1] ? 0 : 1;
    }
  }
  for (int i = 0; i < n; i++)
    if (ans[i] == '?') {
      ans[i] = 'A';
      while (ans[i] == ans[i + 1] || (i > 0 && ans[i] == ans[i - 1])) ans[i]++;
    }
  printf("%s\n", ans);
}