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 inf = 1e09 + 31;
int n, k;
vector<int> a;
const int mxn = 5e5;
int memo[mxn + 5][27];
int dp(int i, int p) {
  if (i == n) return memo[i][p] = (p == a[n - 1]) ? 0 : 0;
  if (~memo[i][p]) return memo[i][p];
  int out = inf;
  for (int j = 0; j < k; j++) {
    if (j == p) continue;
    out = min(out, dp(i + 1, j) + (a[i] != j));
  }
  return memo[i][p] = out;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  memset(memo, -1, sizeof(memo));
  cin >> n >> k;
  string s;
  cin >> s;
  for (int i = 0; i < n; i++) {
    a.push_back(s[i] - 'A');
  }
  int ans = dp(0, 26);
  cout << ans << '\n';
  int prev = 26;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      if (prev != j && memo[i][j] + (j != a[i - 1]) == (ans)) {
        cout << (char)(j + 'A');
        ans = memo[i][j];
        prev = j;
        break;
      }
    }
  }
}