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 oo = 1e9;
const int MAX = 100;
int n, a[4][MAX][MAX], b[MAX * 2][MAX * 2], idx[4], res, sum[2], flag;
char s[MAX + 1];
int main() {
  scanf("%d", &n);
  for (int tt = 0; tt < 4; ++tt)
    for (int i = 0; i < n; ++i) {
      scanf("%s", s);
      for (int j = 0; j < n; ++j) a[tt][i][j] = s[j] - '0';
    }
  for (int i = 0; i < 4; ++i) idx[i] = i;
  res = oo;
  do {
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j) b[i][j] = a[idx[0]][i][j];
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j) b[i][j + n] = a[idx[1]][i][j];
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j) b[i + n][j] = a[idx[2]][i][j];
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j) b[i + n][j + n] = a[idx[3]][i][j];
    sum[0] = sum[1] = 0;
    for (int i = 0; i < n * 2; ++i)
      for (int j = 0; j < n * 2; ++j) {
        flag = (i + j) & 1;
        sum[0] += (b[i][j] != flag);
        sum[1] += (b[i][j] != (1 - flag));
      }
    res = min(res, sum[0]);
    res = min(res, sum[1]);
  } while (next_permutation(idx, idx + 4));
  printf("%d", res);
  return 0;
}
