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 n, k, mem[500010][28];
string s, f;
int dp(int i = 0, int j = 27) {
  if (i == n) return 0;
  int& ret = mem[i][j];
  if (~ret) return ret;
  int c1 = 1e9, c2 = 1e9;
  if (j != s[i] - 'A') c1 = dp(i + 1, s[i] - 'A');
  for (int u = 0; u < min(3, k); u++) {
    if (j != u && u != s[i] - 'A') c2 = min(c2, dp(i + 1, u) + 1);
  }
  return ret = min(c1, c2);
}
void dp2(int i = 0, int j = 27) {
  if (i == n) return;
  if (j != s[i] - 'A' && dp(i, j) == dp(i + 1, s[i] - 'A')) {
    cout << s[i];
    dp2(i + 1, s[i] - 'A');
  } else {
    for (int u = 0; u < min(3, k); u++) {
      if (j != u && u != s[i] - 'A' && dp(i, j) == dp(i + 1, u) + 1) {
        cout << (char)('A' + u);
        dp2(i + 1, u);
        break;
      }
    }
  }
}
int main() {
  cin >> n >> k;
  cin >> s;
  memset(mem, -1, sizeof mem);
  cout << dp() << endl;
  dp2();
  return 0;
}