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[500010];
int n, k, dp[500010][26];
void dfs(int id, int d, int pre) {
  if (id < 0) return;
  for (int i = 0; i < k; i++)
    if (dp[id][i] == d && i != pre) {
      dfs(id - 1, d - (i != s[id]), i);
      putchar(i + 'A');
      break;
    }
}
int main() {
  while (scanf("%d%d", &n, &k) == 2) {
    scanf("%s", s);
    for (int i = 0; i < n; i++) s[i] -= 'A';
    for (int i = 1; i < n; i++)
      for (int j = 0; j < k; j++) dp[i][j] = 0x3fffffff;
    for (int j = 0; j < k; j++) dp[0][j] = (j != s[0]);
    for (int i = 1; i < n; i++)
      for (int j = 0; j < k; j++)
        for (int r = 0; r < k; r++)
          if (j != r) dp[i][j] = min(dp[i][j], dp[i - 1][r] + (s[i] != j));
    int Min = dp[n - 1][0];
    for (int i = 1; i < k; i++) Min = min(Min, dp[n - 1][i]);
    printf("%d\n", Min);
    dfs(n - 1, Min, -1);
    puts("");
  }
  return 0;
}