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 MAXN = 128;
bool a[4][MAXN][MAXN];
long long cnt[4][2];
long long ans;
string s = "0123";
long long n;
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cerr << "HELLO WORLD :)\n";
  cin >> n;
  for (long long i = 0; i < 4; i++) {
    for (long long j = 0; j < n; j++) {
      string rel;
      cin >> rel;
      for (long long k = 0; k < n; k++) {
        a[i][j][k] = rel[k] - '0';
      }
    }
  }
  for (long long i = 0; i < 4; i++) {
    for (long long c = 0; c < 2; c++) {
      for (long long j = 0; j < n; j++) {
        for (long long k = 0; k < n; k++) {
          if (!c) {
            if (a[i][j][k] && (j + k) % 2 == 0) cnt[i][c]++;
            if (!a[i][j][k] && (j + k) % 2) cnt[i][c]++;
          } else {
            if (!a[i][j][k] && (j + k) % 2 == 0) cnt[i][c]++;
            if (a[i][j][k] && (j + k) % 2) cnt[i][c]++;
          }
        }
      }
    }
  }
  ans = 1e8;
  do {
    long long x1 = s[0] - '0', x2 = s[1] - '0', x3 = s[2] - '0',
              x4 = s[3] - '0';
    ans = min(ans, cnt[x1][0] + cnt[x2][1] + cnt[x3][1] + cnt[x4][0]);
    ans = min(ans, cnt[x1][1] + cnt[x2][0] + cnt[x3][0] + cnt[x4][1]);
  } while (next_permutation(s.begin(), s.end()));
  return cout << ans << '\n', false;
}
