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 = 105;
int n, res = 1e9, a[5][N][N];
void calc(int a[N][N], int b[N][N], int c[N][N], int d[N][N]) {
  int tmp = 0, tmp2 = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) {
      if (a[i][j] != (i + j) % 2)
        ++tmp;
      else
        ++tmp2;
      if (c[i][j] != (i + j) % 2)
        ++tmp;
      else
        ++tmp2;
      if (b[i][j] != (i + j) % 2)
        ++tmp2;
      else
        ++tmp;
      if (d[i][j] != (i + j) % 2)
        ++tmp2;
      else
        ++tmp;
    }
  res = min(res, min(tmp, tmp2));
}
int main() {
  cin >> n;
  for (int i = 1; i <= 4; ++i)
    for (int j = 1; j <= n; ++j)
      for (int k = 1; k <= n; ++k) {
        char z;
        cin >> z;
        a[i][j][k] = (int)(z - '0');
      }
  vector<int> v;
  v.push_back(1);
  v.push_back(2);
  v.push_back(3);
  v.push_back(4);
  do {
    calc(a[v[0]], a[v[1]], a[v[2]], a[v[3]]);
  } while (next_permutation(v.begin(), v.end()));
  cout << res << "\n";
}
