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;
using lli = long long int;
const int MOD = 1000000007;
const int MOD1 = 998244353;
const int maxn = 100010;
const int lim = (int)1e9;
int aux[5 * maxn][26];
int dp[5 * maxn][26];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n = 0, k = 0;
  string s;
  cin >> n >> k;
  cin >> s;
  for (int i = 0; i < 26; ++i) {
    aux[0][i] = -1;
    dp[0][i] = 1;
  }
  dp[0][s[0] - 'A'] = 0;
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j < k; ++j) {
      aux[i][j] = (j ? 0 : 1);
      dp[i][j] = dp[i - 1][j ? 0 : 1] + 1;
      for (int l = 1; l < k; ++l)
        if (l != j && dp[i][j] > dp[i - 1][l] + 1) {
          aux[i][j] = l;
          dp[i][j] = dp[i - 1][l] + 1;
        }
    }
    --dp[i][s[i] - 'A'];
  }
  int pos = 0;
  for (int i = 0; i < k; ++i)
    if (dp[n - 1][pos] > dp[n - 1][i]) pos = i;
  cout << dp[n - 1][pos] << endl;
  int pt = n - 1;
  while (pt >= 0) {
    s[pt] = pos + 'A';
    pos = aux[pt][pos];
    --pt;
  }
  cout << s;
}