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;
template <class T>
using v2d = vector<vector<T>>;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
const int maxN = 1e2 + 10;
int n, a[4][maxN][maxN], ans = 1e9;
;
void cal(int board, pair<int, int> &ans) {
  ans.first = ans.second = 0;
  for (int i = 1; i <= (int)(n); ++i) {
    for (int j = 1; j <= (int)(n); ++j) {
      if ((i + j) % 2 == 0) {
        ans.first += (a[board][i][j] != 0);
      } else {
        ans.first += (a[board][i][j] != 1);
      }
    }
  }
  for (int i = 1; i <= (int)(n); ++i) {
    for (int j = 1; j <= (int)(n); ++j) {
      if ((i + j) % 2 == 0) {
        ans.second += (a[board][i][j] != 1);
      } else {
        ans.second += (a[board][i][j] != 0);
      }
    }
  }
}
void solve() {
  cin >> n;
  for (int board = 0; board < (int)(4); ++board) {
    for (int i = 1; i <= (int)(n); ++i) {
      for (int j = 1; j <= (int)(n); ++j) {
        char c;
        cin >> c;
        a[board][i][j] = c - '0';
      }
    }
  }
  vector<pair<int, int>> vec(4);
  for (int board = 0; board < (int)(4); ++board) {
    cal(board, vec[board]);
    cerr << vec[board].first << ' ' << vec[board].second << '\n';
  }
  vector<int> per(4);
  iota((per).begin(), (per).end(), 0);
  for (int iter = 1; iter <= (int)(24); ++iter) {
    int cur = vec[per[0]].first + vec[per[1]].first + vec[per[2]].second +
              vec[per[3]].second;
    uin(ans, cur);
    next_permutation((per).begin(), (per).end());
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
