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 int N = 502;
int n, a[5][N][N];
int ans = 1000000000;
int bst1[N][N], bst2[N][N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 4; ++i)
    for (int j = 1; j <= n; ++j)
      for (int k = 1; k <= n; ++k) scanf("%1d", &a[i][j][k]);
  for (int i = 1; i <= 2 * n; ++i)
    for (int j = 1; j <= 2 * n; ++j) {
      bst1[i][j] = (i + j) & 1;
      bst2[i][j] = (i + j) % 2 == 0;
    }
  int p[] = {1, 2, 3, 4};
  do {
    int curr = 0;
    for (int i = 0; i <= 3; ++i) {
      for (int x = 1; x <= n; ++x)
        for (int y = 1; y <= n; ++y)
          curr += a[p[i]][x][y] != bst1[x + n * (i > 1)][y + n * (i & 1)];
    }
    ans = min(ans, curr);
    curr = 0;
    for (int i = 0; i <= 3; ++i) {
      for (int x = 1; x <= n; ++x)
        for (int y = 1; y <= n; ++y)
          curr += a[p[i]][x][y] != bst2[x + n * (i > 1)][y + n * (i & 1)];
    }
    ans = min(ans, curr);
  } while (next_permutation(p, p + 4));
  cout << ans;
}
