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;
vector<pair<int, int> > t;
int main() {
  string s;
  t.resize(4);
  int n;
  cin >> n;
  for (int k = 0; k < 4; k++) {
    int ans = 0;
    for (int i = 0; i < n; i++) {
      cin >> s;
      for (int j = 0; j < s.size(); j++)
        if ((i * n + j) % 2 == s[j] - '0') {
          ans++;
        }
    }
    t[k].first = ans;
    t[k].second = n * n - ans;
  }
  sort(t.begin(), t.end());
  long long s2 = (long long)t[0].first + (long long)t[1].first +
                 (long long)t[2].second + (long long)t[3].second;
  long long res = s2;
  do {
    s2 = (long long)t[0].first + (long long)t[1].first +
         (long long)t[2].second + (long long)t[3].second;
    res = min(s2, res);
  } while (next_permutation(t.begin(), t.end()));
  s2 = (long long)t[0].first + (long long)t[1].first + (long long)t[2].second +
       (long long)t[3].second;
  res = min(s2, res);
  cout << res << endl;
  return 0;
}
