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;
inline int read() {
  int x = 0, neg = 1;
  char c = getchar();
  while (c < 48 || c > 57) {
    if (c == 45) neg = -neg;
    c = getchar();
  }
  while (c >= 48 && c <= 57) {
    x = (x << 1) + (x << 3) + c - 48;
    c = getchar();
  }
  return x * neg;
}
inline void no() { cout << "NO\n"; }
inline void yes() { cout << "YES\n"; }
const int omg = 1e9, mod = 1e9 + 7;
void solve() {
  int N;
  cin >> N;
  int arr[4] = {0};
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < N; j++) {
      for (int k = 0; k < N; k++) {
        char c;
        cin >> c;
        if ((c == '0' && k % 2 == j % 2) || c == '1' && k % 2 != j % 2)
          arr[i]++;
      }
    }
  }
  sort(arr, arr + 4);
  cout << arr[0] + arr[1] - arr[2] - arr[3] + 2 * pow(N, 2);
}
signed main() {
  solve();
  return 0;
}
