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;
char arr[4][105][105];
int w[2][4];
int p[4];
int ans;
int n;
int calc(int id, int v) {
  int ret = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (arr[id][i][j] != v + '0') {
        ++ret;
      }
      v ^= 1;
    }
  }
  return ret;
}
void solve() {
  scanf("%d", &n);
  for (int k = 0; k < 4; ++k) {
    for (int i = 0; i < n; ++i) {
      scanf("%s", arr[k][i]);
    }
  }
  for (int k = 0; k < 4; ++k) {
    for (int i = 0; i < 2; ++i) {
      w[i][k] = calc(k, i);
    }
  }
  for (int i = 0; i < 4; ++i) {
    p[i] = i;
  }
  ans = 1e9;
  do {
    for (int i = 0; i < 2; ++i) {
      int cur = 0;
      for (int j = 0; j < 4; ++j) {
        cur += w[(j == 0 || j == 3) ^ i][p[j]];
      }
      ans = min(ans, cur);
    }
  } while (next_permutation(p, p + 4));
  printf("%d\n", ans);
}
int main() {
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}
