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_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, ans = 1000000007;
  cin >> n;
  int a[2][2 * n][2 * n], b[4][n][n], c[2 * n][2 * n];
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2 * n; j++) {
      for (int k = 0; k < 2 * n; k++) {
        a[i][j][k] = (i + j + k) % 2;
      }
    }
  }
  string s;
  vector<int> v;
  for (int i = 0; i < 4; i++) {
    v.push_back(i);
    for (int k = 0; k < n; k++) {
      cin >> s;
      for (int j = 0; j < n; j++) {
        b[i][k][j] = s[j] - '0';
      }
    }
  }
  do {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        c[i][j] = b[v[0]][i][j];
        c[i][j + n] = b[v[1]][i][j];
        c[i + n][j] = b[v[2]][i][j];
        c[i + n][j + n] = b[v[3]][i][j];
      }
    }
    int c1 = 0, c2 = 0;
    for (int i = 0; i < 2 * n; i++) {
      for (int j = 0; j < 2 * n; j++) {
        if (c[i][j] != a[0][i][j]) c1++;
        if (c[i][j] != a[1][i][j]) c2++;
      }
    }
    ans = min(ans, min(c1, c2));
  } while (next_permutation(v.begin(), v.end()));
  cout << ans;
}
