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 long long INF = 1e9;
const long long MAXN = 1e2 + 69;
const long long MOD = 1e9 + 7;
int n, ans, cnt[2], res[4][2];
bool mark[4];
string arr[MAXN];
vector<pair<pair<int, int>, int> > vect;
inline int check(int type) {
  int cnt = n * n;
  if (type)
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) {
        if (i % 2 and j % 2 and arr[i][j] == '1' or
            !(i % 2) and !(j % 2) and arr[i][j] == '1')
          cnt--;
        if (!(i % 2) and j % 2 and arr[i][j] == '0' or
            i % 2 and !(j % 2) and arr[i][j] == '0')
          cnt--;
      }
  else
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) {
        if (!(i % 2) and !(j % 2) and arr[i][j] == '0' or
            i % 2 and j % 2 and arr[i][j] == '0')
          cnt--;
        if (i % 2 and !(j % 2) and arr[i][j] == '1' or
            !(i % 2) and j % 2 and arr[i][j] == '1')
          cnt--;
      }
  return (n * n) - cnt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int k = 0; k < 4; k++) {
    for (int i = 0; i < n; i++) cin >> arr[i];
    res[k][0] = check(0), res[k][1] = check(1);
  }
  for (int i = 0; i < 4; i++)
    for (int j = 0; j < 2; j++)
      vect.push_back(make_pair(make_pair(res[i][j], j), i));
  sort(vect.begin(), vect.end());
  for (int i = 0; i < vect.size(); i++)
    if (!mark[vect[i].second]) {
      if (cnt[vect[i].first.second] < 2)
        ans += vect[i].first.first, cnt[vect[i].first.second]++;
    }
  cout << ans;
  return cout << '\n', 0;
}
