Magnus decided to play a classic chess game. Though what he saw in his locker shocked him! His favourite chessboard got broken into 4 pieces, each of size n by n, n is always odd. And what's even worse, some squares were of wrong color. j-th square of the i-th row of k-th piece of the board has color ak, i, j; 1 being black and 0 being white. 

Now Magnus wants to change color of some squares in such a way that he recolors minimum number of squares and obtained pieces form a valid chessboard. Every square has its color different to each of the neightbouring by side squares in a valid board. Its size should be 2n by 2n. You are allowed to move pieces but not allowed to rotate or flip them.

Input

The first line contains odd integer n (1 ≤ n ≤ 100) — the size of all pieces of the board. 

Then 4 segments follow, each describes one piece of the board. Each consists of n lines of n characters; j-th one of i-th line is equal to 1 if the square is black initially and 0 otherwise. Segments are separated by an empty line.

Output

Print one number — minimum number of squares Magnus should recolor to be able to obtain a valid chessboard.

Examples

Input

1
0

0

1

0


Output

1


Input

3
101
010
101

101
000
101

010
101
011

010
101
010


Output

2
#include <bits/stdc++.h>
using namespace std;
const long long aN = 2e5 + 5;
const long long mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  int tc = 1;
  while (tc--) {
    long long i, j, k, w, b, n, out;
    cin >> n;
    vector<pair<int, int> > vp;
    vector<string> v(n);
    for (k = 0; k < 4; k++) {
      w = 0;
      b = 0;
      for (i = 0; i < n; i++) cin >> v[i];
      for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
          if ((i + j) % 2 == 0 && v[i][j] == '1') w++;
          if ((i + j) % 2 == 0 && v[i][j] == '0') b++;
          if ((i + j) % 2 == 1 && v[i][j] == '1') b++;
          if ((i + j) % 2 == 1 && v[i][j] == '0') w++;
        }
      }
      vp.push_back(make_pair(w, b));
    }
    sort(vp.begin(), vp.end());
    out = vp[0].first + vp[1].first + vp[2].second + vp[3].second;
    cout << out << endl;
  }
  return 0;
}
