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 dp[511111];
char s[511111];
char s1[511111];
int n;
void ans() {
  int ans1 = 0;
  for (int i = 0; i < n; i++) {
    if (i % 2) {
      s1[i] = 'A';
    } else
      s1[i] = 'B';
    if (s1[i] != s[i]) ans1++;
  }
  if (n - ans1 < ans1) {
    cout << n - ans1 << endl;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0)
        cout << 'A';
      else
        cout << 'B';
    }
    cout << endl;
  } else {
    cout << ans1 << endl;
    cout << s1 << endl;
  }
}
int main() {
  memset(dp, 0, sizeof(dp));
  int k;
  int sta;
  int start = 0, end = 0, max = 0;
  cin >> n >> k;
  cin >> s;
  if (k == 2) {
    ans();
  } else {
    for (int i = 1; i < n; i++) {
      if (s[i] == s[i - 1]) {
        dp[i] = 0;
        start = i;
      } else {
        dp[i] += dp[i - 1] + 1;
        if (max < dp[i]) {
          max = dp[i];
          sta = start;
          end = i;
        }
      }
    }
    if (max == 0) {
      sta = end = 0;
    }
    int ans = 0;
    for (int i = sta - 1; i >= 0; i--) {
      if (s[i] == s[i + 1]) {
        int t = 1;
        while (t + 'A' - 1 == s[i + 1] || t + 'A' - 1 == s[i - 1]) {
          t++;
        }
        if (t > k) {
          t = 1;
          while (t + 'A' - 1 == s[i + 1]) {
            t++;
          }
        }
        s[i] = t + 'A' - 1;
        ans++;
      }
    }
    for (int i = end + 1; i < n; i++) {
      if (s[i] == s[i - 1]) {
        int t = 1;
        while (t + 'A' - 1 == s[i - 1] || t + 'A' - 1 == s[i + 1]) {
          t++;
        }
        if (t > k) {
          t = 1;
          while (t + 'A' - 1 == s[i - 1]) {
            t++;
          }
        }
        s[i] = t + 'A' - 1;
        ans++;
      }
    }
    cout << ans << endl;
    cout << s << endl;
  }
  return 0;
}