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 = 100;
const int M = 4;
const int INF = N * N * 4;
int n;
char c[N + 10];
int a[M][N][N];
int m[M][2];
int main() {
  scanf("%d ", &n);
  for (int(i) = (0); (i) < (int)(M); (i)++) {
    for (int(j) = (0); (j) < (int)(n); (j)++) {
      fgets(c, n + 2, stdin);
      for (int(k) = (0); (k) < (int)(n); (k)++) a[i][j][k] = c[k] - '0';
    }
    fgets(c, 2, stdin);
  }
  memset(m, 0, sizeof(m));
  for (int(i) = (0); (i) < (int)(M); (i)++)
    for (int(j) = (0); (j) < (int)(2); (j)++) {
      for (int(k) = (0); (k) < (int)(n); (k)++)
        for (int(l) = (0); (l) < (int)(n); (l)++) {
          int g = (k + l) & 1 ? (j + 1) & 1 : j;
          if (a[i][k][l] != g) m[i][j]++;
        }
    }
  int res = INF;
  for (int(i) = (0); (i) < (int)(1 << M); (i)++) {
    int cnt = 0;
    for (int(j) = (0); (j) < (int)(M); (j)++)
      if ((i >> j) & 1) cnt++;
    if (cnt != 2) continue;
    cnt = 0;
    for (int(j) = (0); (j) < (int)(M); (j)++) cnt += m[j][(i >> j) & 1];
    res = min(res, cnt);
  }
  printf("%d\n", res);
  return 0;
}
