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;
void solve() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, k;
  cin >> n >> k;
  string second;
  cin >> second;
  if (k > 2) {
    int c = 0;
    for (int i = 1; i < n; i++) {
      if (second[i] == second[i - 1]) {
        if (i + 1 < n) {
          int j = 0;
          while (j < k &&
                 ((j + 'A') == second[i - 1] || (j + 'A') == second[i + 1]))
            j++;
          second[i] = j + 'A';
        } else {
          second[i] = ((second[i - 1] - 'A' + 1) % k) + 'A';
        }
        c++;
      }
    }
    cout << c << '\n';
    cout << second << '\n';
    return;
  }
  int dp[n][k];
  vector<multiset<int>> M(n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < k; j++) dp[i][j] = INT_MAX;
  }
  for (int i = 0; i < k; i++) {
    if (i != (second[0] - 'A')) {
      dp[0][i] = 1;
      M[0].insert(1);
    } else {
      dp[0][i] = 0;
      M[0].insert(0);
    }
  }
  for (int i = 1; i < n; i++) {
    int c = second[i] - 'A';
    for (int j = 0; j < k; j++) {
      int x = dp[i - 1][j];
      M[i - 1].erase(M[i - 1].find(x));
      int y = (*(M[i - 1].begin()));
      M[i - 1].insert(x);
      if (j == c)
        dp[i][j] = min(dp[i][j], y);
      else
        dp[i][j] = min(dp[i][j], y + 1);
      M[i].insert(dp[i][j]);
    }
    M[i - 1].clear();
  }
  int ans = INT_MAX;
  int c;
  for (int i = 0; i < k; i++) {
    if (dp[n - 1][i] < ans) {
      ans = dp[n - 1][i];
      c = i;
    }
  }
  stack<int> st;
  int l = INT_MAX;
  int j = -1;
  for (int i = 0; i < k; i++) {
    if (l > dp[n - 1][i]) {
      l = dp[n - 1][i];
      j = i;
    }
  }
  st.push(j);
  for (int i = n - 2; i >= 0; i--) {
    int ok;
    if (j == (second[i + 1] - 'A'))
      ok = 0;
    else
      ok = -1;
    for (int h = 0; h < k; h++) {
      if (dp[i][h] == (l + ok) && h != j) {
        l = dp[i][h];
        j = h;
        break;
      }
    }
    st.push(j);
  }
  cout << ans << '\n';
  while (!st.empty()) {
    cout << (char)(st.top() + 'A');
    st.pop();
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}