Problem: For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.

Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.

Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.

Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.

Output

Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.

It can be shown that a solution always exists under the given constraints.

Examples

Input

3
999999 0
0 999999
999999 0


Output

1 1 -1 


Input

1
-824590 246031


Output

1 


Input

8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899


Output

1 1 1 1 1 1 1 -1 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct inst {
  int x;
  int y;
  int z;
};
inst ax[100010];
int bn[100010];
int main(void) {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &ax[i].x, &ax[i].y);
    ax[i].z = i;
  }
  while (1) {
    random_shuffle(ax, ax + n);
    long long posx = ax[0].x;
    long long posy = ax[0].y;
    bn[ax[0].z] = 1;
    for (int i = 1; i < n; i++) {
      if ((ax[i].x + posx) * (ax[i].x + posx) +
              (ax[i].y + posy) * (ax[i].y + posy) <=
          (ax[i].x - posx) * (ax[i].x - posx) +
              (ax[i].y - posy) * (ax[i].y - posy)) {
        posx += ax[i].x;
        posy += ax[i].y;
        bn[ax[i].z] = 1;
      } else {
        posx -= ax[i].x;
        posy -= ax[i].y;
        bn[ax[i].z] = -1;
      }
    }
    if (posx * posx + posy * posy <= (long long)1500000 * (long long)1500000)
      break;
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", bn[i]);
  }
  printf("\n");
  return 0;
}