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;
struct P {
  int x, y, z;
};
vector<int> v[55];
int a, c, i, b, k, n, d, m;
int l[255111];
int o[100055];
int dx[10] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[10] = {1, 0, -1, 0, -1, 1, -1, 1},
    dz[10] = {0, 0, 0, 0, 1, -1};
int e;
long long x, y, mod = 1000000007;
long long z;
int j[400200];
P u[101000];
stack<int> s;
queue<int> q;
char r[5][111][111];
bool as(P a, P b) { return a.y < b.y; }
int main() {
  scanf("%d", &a);
  for (int t = 1; t <= 4; t++) {
    for (int w = 1; w <= a; w++) scanf("%s", r[t][w] + 1);
  }
  m = 987654321;
  for (int t = 1; t < 4; t++)
    for (int w = t + 1; w <= 4; w++) {
      n = 0;
      for (int h = 1; h <= 4; h++)
        if (h == t || h == w)
          for (int i = 1; i <= a; i++)
            for (int j = 1; j <= a; j++)
              n += (i + j + r[h][i][j] - '0' + 1) % 2;
      for (int h = 1; h <= 4; h++)
        if (h != t && h != w)
          for (int i = 1; i <= a; i++)
            for (int j = 1; j <= a; j++) n += (i + j + r[h][i][j] - '0') % 2;
      m = min(m, n);
    }
  printf("%d", m);
}
