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;
void open() { freopen("data.txt", "r", stdin); }
void out() { freopen("out.txt", "w", stdout); }
const int maxn = 1e5 + 5;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int pos[5];
int maps[5][105][105], tmp1[205][205], tmp2[205][205];
int main() {
  for (int i = (0); i < (5); ++i) pos[i] = i;
  int n, ans = INF;
  scanf("%d", &n);
  for (int k = (1); k < (5); ++k)
    for (int i = (1); i < (n + 1); ++i)
      for (int j = (1); j < (n + 1); ++j) scanf("%1d", &maps[k][i][j]);
  do {
    for (int i = (1); i < (n + 1); ++i)
      for (int j = (1); j < (n + 1); ++j) {
        tmp1[i][j] = maps[pos[1]][i][j];
        tmp1[i][j + n] = maps[pos[2]][i][j];
        tmp1[i + n][j] = maps[pos[3]][i][j];
        tmp1[i + n][j + n] = maps[pos[4]][i][j];
      }
    int cnt = 0;
    tmp2[1][1] = tmp1[1][1];
    for (int j = (2); j < (2 * n + 1); ++j) tmp2[1][j] = tmp2[1][j - 1] ^ 1;
    for (int i = (2); i < (2 * n + 1); ++i)
      for (int j = (1); j < (2 * n + 1); ++j) tmp2[i][j] = tmp2[i - 1][j] ^ 1;
    for (int i = (1); i < (2 * n + 1); ++i)
      for (int j = (1); j < (2 * n + 1); ++j) cnt += tmp1[i][j] ^ tmp2[i][j];
    cnt = min(cnt, 4 * n * n - cnt);
    ans = min(ans, cnt);
  } while (next_permutation(pos + 1, pos + 5));
  printf("%d\n", ans);
}
