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 ele {
  long long x;
  long long y;
  int index;
};
int wynik[100010];
bool comp(ele a, ele b) {
  if (a.y * b.x - a.x * b.y < 0) return true;
  return false;
}
long long xd = 2250000000000;
ele dane[100010];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, i, in, x, y;
  cin >> n;
  long long totx = 0;
  long long toty = 0;
  for (i = 0; i < n; i++) {
    cin >> x >> y;
    dane[i].x = x;
    dane[i].y = y;
    dane[i].index = i;
  }
  sort(dane, dane + n, comp);
  for (i = 0; i < n; i++) {
    x = dane[i].x;
    y = dane[i].y;
    if (abs((totx + x) * (totx + x) + (toty + y) * (toty + y)) <
        abs((totx - x) * (totx - x) + (toty - y) * (toty - y))) {
      totx += x;
      toty += y;
      wynik[dane[i].index] = 1;
    } else {
      totx -= x;
      toty -= y;
      wynik[dane[i].index] = -1;
    }
  }
  for (i = 0; i < n; i++) {
    cout << wynik[i] << " ";
  }
}