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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  long long n, flag = 0;
  cin >> n;
  std::vector<pair<long double, long double> > v(n);
  for (int i = 0; i < n; i++) cin >> v[i].first >> v[i].second;
  std::vector<long long> ans(n);
  long double px = v[0].first, py = v[0].second;
  ans[0] = 1;
  for (int i = 1; i < n; i++) {
    long double addx = (px + v[i].first);
    long double addy = (py + v[i].second);
    long double subx = (px - v[i].first);
    long double suby = (py - v[i].second);
    if (addx * addx + addy * addy > subx * subx + suby * suby) {
      ans[i] = -1;
      px -= v[i].first;
      py -= v[i].second;
    } else {
      ans[i] = 1;
      px += v[i].first;
      py += v[i].second;
    }
  }
  long double c = 1500000;
  if (px * px + py * py > c * c) {
    ans[0] = -1;
    px = -v[0].first;
    py = -v[0].second;
    for (int i = 1; i < n; i++) {
      long double addx = (px + v[i].first);
      long double addy = (py + v[i].second);
      long double subx = (px - v[i].first);
      long double suby = (py - v[i].second);
      if (addx * addx + addy * addy > subx * subx + suby * suby) {
        ans[i] = -1;
        px -= v[i].first;
        py -= v[i].second;
      } else {
        ans[i] = 1;
        px += v[i].first;
        py += v[i].second;
      }
    }
  }
  if (px * px + py * py > c * c) {
    ans[n - 1] = 1;
    px = -v[n - 1].first;
    py = -v[n - 1].second;
    for (int i = n - 2; i > -1; i--) {
      long double addx = (px + v[i].first);
      long double addy = (py + v[i].second);
      long double subx = (px - v[i].first);
      long double suby = (py - v[i].second);
      if (addx * addx + addy * addy > subx * subx + suby * suby) {
        ans[i] = 1;
        px -= v[i].first;
        py -= v[i].second;
      } else {
        ans[i] = -1;
        px += v[i].first;
        py += v[i].second;
      }
    }
  }
  if (px * px + py * py > c * c) {
    ans[n - 1] = -1;
    px += v[n - 1].first;
    py += v[n - 1].second;
    for (int i = n - 2; i > -1; i--) {
      long double addx = (px + v[i].first);
      long double addy = (py + v[i].second);
      long double subx = (px - v[i].first);
      long double suby = (py - v[i].second);
      if (addx * addx + addy * addy > subx * subx + suby * suby) {
        ans[i] = 1;
        px -= v[i].first;
        py -= v[i].second;
      } else {
        ans[i] = -1;
        px += v[i].first;
        py += v[i].second;
      }
    }
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}