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 long long N = 1e3 + 69, base = 1024 * 1024, mod = 1e9 + 7;
int t[5][N][N], p[N][N], n;
long long test(vector<int> vek) {
  int x[] = {0, 0, n, n};
  int y[] = {0, n, 0, n};
  for (int k = 0; k < 4; k++) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        p[i + x[k]][j + y[k]] = t[vek[k]][i][j];
      }
    }
  }
  int re1 = 0, re2 = 0;
  for (int i = 0; i < 2 * n; i++) {
    for (int j = 0; j < 2 * n; j++) {
      if (p[i][j] == (i + j) % 2)
        re1++;
      else
        re2++;
    }
  }
  return min(re1, re2);
}
int32_t main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int k = 1; k <= 4; k++) {
    for (int i = 0; i < n; i++) {
      string S;
      cin >> S;
      for (int j = 0; j < n; j++) {
        t[k][i][j] = S[j] - '0';
      }
    }
  }
  vector<int> perm;
  for (int i = 1; i <= 4; i++) perm.push_back(i);
  long long wynik = test(perm);
  while (next_permutation(perm.begin(), perm.end()))
    wynik = min(wynik, test(perm));
  cout << wynik;
}
