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 main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<pair<int, int>> need(4);
  for (int t = 0; t < 4; t++) {
    char board[110][110];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        cin >> board[i][j];
      }
    }
    int black = 0;
    int moves = 0;
    for (int i = 0; i < n; i++) {
      black = !black;
      for (int j = 0; j < n; j++) {
        if (j % 2 == 0) {
          if (board[i][j] - '0' != black) {
            moves++;
          }
        } else {
          if (board[i][j] - '0' == black) {
            moves++;
          }
        }
      }
    }
    need[t].first = moves;
    need[t].second = n * n - moves;
  }
  int ans = 0;
  sort(need.begin(), need.end());
  ans += need[0].first;
  ans += need[1].first;
  ans += need[2].second;
  ans += need[3].second;
  cout << ans << endl;
  return 0;
}
