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;
bool D[200][200][2];
bool I[100][100][4];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < 2 * n; i++) {
    for (int j = 0; j < 2 * n; j++) {
      D[i][j][0] = (i + j) % 2;
      D[i][j][1] = (i + j + 1) % 2;
    }
  }
  for (int k = 0; k < 4; k++) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        char c;
        cin >> c;
        I[i][j][k] = (bool)(c - '0');
      }
    }
  }
  int ans = n * n * 40;
  int C[] = {0, 1, 2, 3};
  do {
    int cur[2] = {0, 0};
    for (int k = 0; k < 4; k++) {
      int modcol = C[k] % 2, modrow = (C[k] >= 2 ? 1 : 0);
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
          cur[0] +=
              (I[i][j][k] == D[i + modcol * n][j + modrow * n][0] ? 0 : 1);
          cur[1] +=
              (I[i][j][k] == D[i + modcol * n][j + modrow * n][1] ? 0 : 1);
        }
      }
    }
    ans = min(ans, min(cur[0], cur[1]));
  } while (next_permutation(C, C + 4));
  cout << ans << endl;
  return 0;
}
