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;
template <typename T, typename U>
std::pair<T, U> operator+(const std::pair<T, U>& l, const std::pair<T, U>& r) {
  return {l.first + r.first, l.second + r.second};
}
typedef void (*callback_function)(void);
const long long INF64 = 1e18;
const long long INF32 = 1e9;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  {
    long long n;
    cin >> n;
    char correct[2][n][n];
    for (long long k = (0); k < (2); k++)
      for (long long i = (0); i < (n); i++)
        for (long long j = (0); j < (n); j++) {
          correct[k][i][j] = char('0' + (i + j + k) % 2);
        }
    long long d[4][2], dist;
    char a[n][n];
    for (long long t = (0); t < (4); t++) {
      for (long long i = (0); i < (n); i++)
        for (long long j = (0); j < (n); j++) cin >> a[i][j];
      for (long long k = (0); k < (2); k++) {
        dist = 0;
        for (long long i = (0); i < (n); i++)
          for (long long j = (0); j < (n); j++)
            dist += (a[i][j] != correct[k][i][j]);
        d[t][k] = dist;
      }
    }
    long long res = INF32, r;
    for (long long i = (0); i < (3); i++)
      for (long long j = (i + 1); j < (4); j++) {
        r = d[i][0] + d[j][0];
        for (long long k = (0); k < (4); k++)
          if (k != i && k != j) r += d[k][1];
        res = min(res, r);
      }
    cout << res << endl;
  }
  return 0;
}
