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 long long mod = (long long)1e9 + 7;
const long long inf = (long long)1e16;
const long double eps = 1e-12;
const long long N = (long long)5e5 + 5;
const long long LOGN = 19;
const long double PI = 3.14159265358979323846;
inline long long mul(long long a, long long b, long long m = mod) {
  return (long long)(a * b) % m;
}
inline long long add(long long a, long long b, long long m = mod) {
  a += b;
  if (a >= m) a -= m;
  if (a < 0) a += m;
  return a;
}
inline long long power(long long a, long long b, long long m = mod) {
  if (b == 0) return 1;
  if (b == 1) return (a % m);
  long long x = power(a, b / 2, m);
  x = mul(x, x, m);
  if (b % 2) x = mul(x, a, m);
  return x;
}
long long n, tt;
string s, ans;
long long dp[N][26], par[N][26];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (0) {
    freopen("C:\\Users\\Dishant\\Desktop\\Collection-DEV c++\\input.txt", "r",
            stdin);
    freopen("C:\\Users\\Dishant\\Desktop\\Collection-DEV c++\\output.txt", "w",
            stdout);
  }
  cin >> n >> tt;
  cin >> s;
  for (long long i = 0; i < tt; i++) {
    if (i == (s[0] - 'A')) continue;
    dp[0][i] = 1;
  }
  for (long long i = 1; i < n; i++) {
    for (long long j = 0; j < tt; j++) {
      long long to = 1;
      if (s[i] - 'A' == j) to = 0;
      long long mn = INT_MAX, id = -1;
      for (long long k = 0; k < tt; k++) {
        if (k == j) continue;
        if (dp[i - 1][k] < mn) {
          mn = dp[i - 1][k];
          id = k;
        }
      }
      dp[i][j] = mn + to;
      par[i][j] = id;
    }
  }
  long long curr = -1, mn = INT_MAX;
  for (long long j = 0; j < tt; j++) {
    if (dp[n - 1][j] < mn) {
      mn = dp[n - 1][j];
      curr = j;
    }
  }
  for (long long i = n - 1; i >= 0; i--) {
    ans += (char)(curr + 'A');
    curr = par[i][curr];
  }
  cout << mn << "\n";
  { reverse(ans.begin(), ans.end()); };
  cout << ans;
  return 0;
}