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;
string A[4][102];
string M[202];
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int p = 0; p < 4; p++) {
    for (int i = 0; i < n; i++) {
      cin >> A[p][i];
    }
  }
  vector<int> T;
  for (int i = 0; i < 4; i++) T.push_back(i);
  int ans = int(1e9);
  do {
    for (int i = 0; i < n; i++) M[i] = A[T[0]][i] + A[T[1]][i];
    for (int i = n; i < 2 * n; i++) M[i] = A[T[2]][i - n] + A[T[3]][i - n];
    int op1 = 0, op2 = 0;
    for (int i = 0; i < 2 * n; i++) {
      int sq = (i % 2) ? 1 : 0;
      for (int j = 0; j < 2 * n; j++) {
        if (M[i][j] - '0' != sq) op1++;
        sq ^= 1;
      }
    }
    for (int i = 0; i < 2 * n; i++) {
      int sq = (i % 2) ? 0 : 1;
      for (int j = 0; j < 2 * n; j++) {
        if (M[i][j] - '0' != sq) op2++;
        sq ^= 1;
      }
    }
    ans = min(ans, min(op1, op2));
  } while (next_permutation(T.begin(), T.end()));
  cout << ans << endl;
}
