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>
#pragma comment(linker, "/STACK:167772160000")
using namespace std;
vector<pair<long long, long long> > tables;
long long oneSum = 0;
long long zeroSum = 0;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < 4; i++) {
    char tmp;
    int missO = 0, missZ = 0;
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        cin >> tmp;
        int sum = j + k;
        if (sum % 2 == 0) {
          missO += (tmp == '0');
          missZ += (tmp == '1');
        } else {
          missO += (tmp == '1');
          missZ += (tmp == '0');
        }
      }
    }
    oneSum += missO;
    zeroSum += missZ;
    tables.push_back(make_pair(missO, missZ));
  }
  long long ans = 999999999999;
  for (int i = 0; i < 4; i++) {
    for (int j = i + 1; j < 4; j++) {
      int miss = tables[i].first + tables[j].first;
      int left = zeroSum - (tables[i].second + tables[j].second);
      ans = min(ans, (long long)miss + (long long)left);
    }
  }
  cout << ans;
  return 0;
}
