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;
int n, p[105][105][4], ans = 1e9;
char c;
int Try(int a, int b, int c, int d) {
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i % 2 == j % 2 && p[i][j][a] == 1) ans++;
      if (i % 2 != j % 2 && p[i][j][a] == 0) ans++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i % 2 == j % 2 && p[i][j][b] == 1) ans++;
      if (i % 2 != j % 2 && p[i][j][b] == 0) ans++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i % 2 == j % 2 && p[i][j][c] == 0) ans++;
      if (i % 2 != j % 2 && p[i][j][c] == 1) ans++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i % 2 == j % 2 && p[i][j][d] == 0) ans++;
      if (i % 2 != j % 2 && p[i][j][d] == 1) ans++;
    }
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int k = 1; k <= 4; k++) {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        cin >> c;
        p[i][j][k] = c - '0';
      }
    }
  }
  ans = min(ans, Try(1, 2, 3, 4));
  ans = min(ans, Try(1, 3, 2, 4));
  ans = min(ans, Try(1, 4, 3, 2));
  ans = min(ans, Try(2, 3, 1, 4));
  ans = min(ans, Try(2, 4, 1, 3));
  ans = min(ans, Try(3, 4, 1, 2));
  cout << ans << endl;
}
