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>
const int N = (int)(1e6) + 322;
const long long INF = (long long)(1e9) + 7;
const double eps = (1e-9);
using namespace std;
long long cnt[N], n, m, ans, t[N], a[N], ans1, mn = INF;
bool ch[1000][1000][5];
vector<int> v;
string s;
long long dx[5] = {0, 1, 1, 0};
long long dy[5] = {0, 0, 1, 1};
vector<string> S;
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  getline(cin, s);
  for (int it = 1; it <= 4; ++it) {
    for (int i = 1; i <= n; ++i) {
      getline(cin, s);
      S.push_back(s);
      for (int j = 1; j <= n; ++j) {
        ch[i][j][it] = (s[j - 1] == '1');
      }
    }
    if (it < 4) getline(cin, s);
    v.push_back(it - 1);
  }
  do {
    ans = ans1 = 0;
    for (int it = 1; it <= 4; ++it) {
      for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
          long long x = dx[v[it - 1]] * n + i;
          long long y = dy[v[it - 1]] * n + j;
          ans += ((x + y) % 2) == ch[i][j][it];
          ans1 += ((x + y) % 2) != ch[i][j][it];
        }
      }
    }
    mn = min(mn, min(ans1, ans));
  } while (next_permutation(v.begin(), v.end()));
  cout << mn;
  return 0;
}
