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;
void JIZZ() {
  cout << "";
  exit(0);
}
const long double PI = 3.14159265358979323846264338327950288;
const long double eps = 1e-13;
const long long mod = 1e9 + 7;
int cl[4][111][111], n;
int cac(int c1[111][111], int c2[111][111], int c3[111][111],
        int c4[111][111]) {
  int rt = 1 << 29, cnt = 0;
  for (int i = 0; i < n * 2; ++i) {
    for (int j = 0; j < 2 * n; ++j) {
      int shou = (i + j) & 1;
      if (i < n && j < n) {
        cnt += c1[i][j] != shou;
      } else if (i < n) {
        cnt += c2[i][j - n] != shou;
      } else if (j < n) {
        cnt += c3[i - n][j] != shou;
      } else
        cnt += c4[i - n][j - n] != shou;
    }
  }
  rt = cnt;
  cnt = 0;
  for (int i = 0; i < n * 2; ++i) {
    for (int j = 0; j < 2 * n; ++j) {
      int shou = (i + j) & 1;
      shou ^= 1;
      if (i < n && j < n) {
        cnt += c1[i][j] != shou;
      } else if (i < n) {
        cnt += c2[i][j - n] != shou;
      } else if (j < n) {
        cnt += c3[i - n][j] != shou;
      } else
        cnt += c4[i - n][j - n] != shou;
      ;
      ;
    }
  };
  ;
  return min(rt, cnt);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int z = 0; z < 4; ++z)
    for (int i = 0; i < n; ++i) {
      string s;
      cin >> s;
      for (int j = 0; j < n; ++j) {
        cl[z][i][j] = (s[j] == '1');
        ;
        ;
      }
    }
  int per[] = {0, 1, 2, 3};
  int ans = 1 << 29;
  do {
    ans = min(ans, cac(cl[per[0]], cl[per[1]], cl[per[2]], cl[per[3]]));
    ;
    ;
  } while (next_permutation(per, per + 4));
  cout << ans << endl;
}
