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 const INF = INT_MAX;
int main() {
  int n, k;
  string s;
  cin >> n >> k >> s;
  string ans_s = s;
  vector<vector<int>> dp(n, vector<int>(k, INF));
  for (int i = 0; i < k; i++) {
    dp[0][i] = 1;
  }
  dp[0][s[0] - 'A'] = 0;
  vector<vector<int>> par(n, vector<int>(k, -1));
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < k; j++) {
      int tmp = INF;
      for (int l = 0; l < k; l++) {
        if (j == l) continue;
        if (dp[i - 1][l] < tmp) {
          par[i][j] = l;
          tmp = dp[i - 1][l];
        }
      }
      dp[i][j] = tmp + ((s[i] - 'A') != j);
    }
  }
  int ans = INF;
  int starter;
  for (int i = 0; i < k; i++) {
    if (dp[n - 1][i] < ans) {
      ans = dp[n - 1][i];
      starter = i;
    }
  }
  cout << ans << '\n';
  string reversed;
  for (int i = n - 1; i >= 0; i--) {
    reversed.push_back((char)('A' + starter));
    starter = par[i][starter];
  }
  reverse(reversed.begin(), reversed.end());
  cout << reversed;
  return 0;
}