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 int maxn = 1e5 + 10;
const int maxm = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int casn, n, m, k;
int num[maxn], vis[maxn], sum[maxn], s[maxn];
int mp[5][500][500];
int t[500][500];
int t2[500][500];
const int dir[9][2] = {1, 0, -1, 0,  0, 1,  0,  -1, 1,
                       1, 1, -1, -1, 1, -1, -1, 0,  0};
int dfs(int x[]) {
  memset(t, -1, sizeof t);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) t[i][j] = mp[x[1]][i][j];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) t[i + n][j] = mp[x[2]][i][j];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) t[i][j + n] = mp[x[3]][i][j];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) t[i + n][j + n] = mp[x[4]][i][j];
  n *= 2;
  int ans2 = 0;
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (t[i][j] != t2[i][j])
        ans++;
      else
        ans2++;
  n /= 2;
  return min(ans, ans2);
}
int main() {
  cin >> n;
  int cnt = 0;
  for (int i = 1; i <= 2 * n; i++) {
    for (int j = 1; j <= 2 * n; j++) {
      t2[i][j] = (cnt++) % 2;
    }
    cnt++;
  }
  for (int k = 1; k <= 4; k++) {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        scanf("%1d", &mp[k][i][j]);
      }
    }
  }
  int ans = INF;
  int pe[] = {0, 1, 2, 3, 4};
  do {
    ans = min(ans, dfs(pe));
  } while (next_permutation(pe + 1, pe + 5));
  cout << ans << endl;
  return 0;
}
