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;
ostream& operator<<(ostream& os, const pair<int, int>& pai) {
  os << pai.first << ' ' << pai.second;
  return os;
}
istream& operator>>(istream& is, pair<int, int>& pai) {
  is >> pai.first >> pai.second;
  return is;
}
int n;
pair<long long, long long> ret;
pair<int, int> arr[100001];
int ans[100001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = n - 1; i >= 0; i--) {
    pair<int, int> t = arr[i];
    long long nx1 = ret.first + t.first, ny1 = ret.second + t.second;
    long long nx2 = ret.first - t.first, ny2 = ret.second - t.second;
    long long d1 = nx1 * nx1 + ny1 * ny1, d2 = nx2 * nx2 + ny2 * ny2;
    if (sqrt(d1 + .0) <= sqrt(d2 + .0)) {
      ans[i] = 1;
      ret = {nx1, ny1};
    } else {
      ans[i] = -1;
      ret = {nx2, ny2};
    }
  }
  for (int i = 0; i < n; i++) cout << ans[i] << ' ';
  return 0;
}