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;
ifstream in("input.txt");
ofstream out("output.txt");
int n, k;
string s, ans;
int dp[500001][30], p[500001][30];
pair<int, int> pref[30], suf[30];
int main() {
  cin >> n >> k;
  cin >> s;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < k; j++) dp[i][j] = 1000000;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < k; j++) {
      if (i == 0) {
        dp[i][j] = 1 - (s[i] == char(j + 'A'));
        continue;
      }
      pair<int, int> opt;
      if (j == 0)
        opt = suf[1];
      else if (j == k - 1)
        opt = pref[j - 1];
      else
        opt = min(pref[j - 1], suf[j + 1]);
      dp[i][j] = opt.first;
      p[i][j] = opt.second;
      if (s[i] != char(j + 'A')) dp[i][j]++;
    }
    pref[0] = make_pair(dp[i][0], 0);
    for (int j = 1; j < k; j++)
      pref[j] = min(pref[j - 1], make_pair(dp[i][j], j));
    suf[k - 1] = make_pair(dp[i][k - 1], k - 1);
    for (int j = k - 2; j >= 0; j--)
      suf[j] = min(suf[j + 1], make_pair(dp[i][j], j));
  }
  int last = -1, minCost = 1000000;
  for (int i = 0; i < k; i++) {
    if (dp[n - 1][i] < minCost) minCost = dp[n - 1][i], last = i;
  }
  for (int i = n - 1; i >= 0; i--) {
    ans += char(last + 'A');
    last = p[i][last];
  }
  reverse(ans.begin(), ans.end());
  cout << minCost << "\n";
  cout << ans;
  return 0;
}