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;
char a[110][110];
int b[10];
int main() {
  int n;
  while (cin >> n) {
    int ans = 0;
    for (int t = 1; t <= 4; t++) {
      int cnt = 0;
      for (int i = 0; i < n; i++) {
        cin >> a[i];
        for (int j = 0; j < n; j++) {
          if (i % 2 == 0 && j % 2 == 0) {
            if (a[i][j] == '0') cnt++;
          }
          if (i % 2 == 0 && j % 2 == 1) {
            if (a[i][j] == '1') cnt++;
          }
          if (i % 2 == 1 && j % 2 == 0) {
            if (a[i][j] == '1') cnt++;
          }
          if (i % 2 == 1 && j % 2 == 1) {
            if (a[i][j] == '0') cnt++;
          }
        }
      }
      b[t] = cnt;
    }
    sort(b + 1, b + 5);
    ans += b[1] + b[2];
    ans += n * n - b[3];
    ans += n * n - b[4];
    cout << ans << endl;
  }
}
