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;
template <typename T>
auto operator<<(ostream& out, T& container)
    -> decltype(container.begin(), container.end(), out) {
  bool first = true;
  out << "[";
  for (auto it = container.begin(); it != container.end(); ++it) {
    if (!first) out << ", ";
    out << *it;
    first = false;
  }
  return out << "]";
}
int main() {
  int n;
  cin >> n;
  int cost[4][2] = {};
  for (int i = 0; i < 4; ++i) {
    for (int r = 0; r < n; ++r)
      for (int c = 0; c < n; ++c) {
        char x;
        cin >> x;
        if (x == '1' == (r % 2 == c % 2))
          cost[i][0]++;
        else
          cost[i][1]++;
      }
  }
  int best = 1ll << 30;
  vector<int> p = {0, 1, 2, 3};
  do {
    best = min(best,
               cost[p[0]][0] + cost[p[1]][0] + cost[p[2]][1] + cost[p[3]][1]);
  } while (next_permutation(p.begin(), p.end()));
  cout << best << endl;
}
