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;
int a[4][2];
void solve() {
  int n;
  cin >> n;
  for (long long int j = 0; j < 4; ++j) {
    string s[n];
    for (long long int i = 0; i < n; ++i) {
      cin >> s[i];
    }
    int o = 0, e = 0;
    for (long long int i = 0; i < n; ++i) {
      for (long long int k = 0; k < n; ++k) {
        if ((i + k) % 2 == 0) {
          if (s[i][k] == '1') e++;
        }
        if ((i + k) % 2 == 1 && s[i][k] == '0') e++;
        if ((i + k) % 2 == 0 && s[i][k] == '0') o++;
        if ((i + k) % 2 == 1 && s[i][k] == '1') o++;
      }
    }
    a[j][0] = o;
    a[j][1] = e;
  }
  int ans = INT_MAX;
  for (int i = 0; i < 3; i++) {
    for (int j = i + 1; j < 4; j++) {
      int cur = 0;
      for (int k = 0; k < 4; k++) {
        if (k == i || k == j)
          cur += a[k][0];
        else
          cur += a[k][1];
      }
      ans = min(ans, cur);
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
  return (0);
}
