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;
const int mod2 = 998244353;
const int mod = 1e9 + 7;
void io() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const int N = 1e6 + 5;
int main() {
  io();
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  int a[n];
  for (int i = 0; i < n; i++) a[i] = s[i] - 'A';
  int dp[n][k];
  int T[n][k];
  for (int j = 0; j < k; j++) {
    dp[0][j] = (j != a[0]);
  }
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < k; j++) {
      dp[i][j] = 1e9;
      for (int l = 0; l < k; l++) {
        if (l == j) continue;
        if (dp[i - 1][l] + (j != a[i]) <= dp[i][j]) {
          dp[i][j] = dp[i - 1][l] + (j != a[i]);
          T[i][j] = l;
        }
      }
    }
  }
  int ans = 1e9;
  int t = 0;
  for (int i = 0; i < k; i++) {
    if (dp[n - 1][i] <= ans) {
      ans = dp[n - 1][i];
      t = i;
    }
  }
  vector<char> res(n + 1);
  res[n] = t;
  for (int i = n - 1; i > 0; i--) {
    t = T[i][t];
    res[i] = t;
  }
  cout << ans << endl;
  for (int i = 1; i <= n; i++) {
    cout << char(res[i] + 'A');
  }
  cout << endl;
}