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;
const int N = 109;
int n;
string a[4][N];
int main() {
  cin >> n;
  for (int k = 0; k < 4; ++k)
    for (int i = 0; i < n; ++i) {
      cin >> a[k][i];
      for (int j = 0; j < n; ++j) a[k][i][j] -= '0';
    }
  vector<int> v;
  for (int k = 0; k < 4; ++k) {
    int sum = 0;
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        if ((i + j) % 2 != a[k][i][j]) ++sum;
    v.push_back(sum);
  }
  int res = int(1e9);
  vector<int> perm;
  for (int k = 0; k < 4; ++k) perm.push_back(k);
  do {
    int sum = 0;
    for (int k = 0; k < 4; ++k) {
      int id = perm[k];
      if (k & 1)
        sum += v[id];
      else
        sum += n * n - v[id];
    }
    res = min(res, sum);
  } while (next_permutation(perm.begin(), perm.end()));
  cout << res << endl;
}
