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 
#include <bits/stdc++.h>
using namespace std;
void Solve() {
  long long n;
  cin >> n;
  vector<pair<pair<long long, long long>, long long> > a(n);
  vector<long long> ans(n);
  for (long long i = 0; i < (n); ++i) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].second = i;
    if (a[i].first.first < 0) {
      a[i].first.first = -a[i].first.first;
      a[i].first.second = -a[i].first.second;
      ans[i] ^= 1;
    }
  }
  sort((a).begin(), (a).end(),
       [](pair<pair<long long, long long>, long long> a,
          pair<pair<long long, long long>, long long> b) {
         return a.first.second * b.first.first < a.first.first * b.first.second;
       });
  long long x = a[0].first.first, y = a[0].first.second;
  for (long long i = 1; i < n; ++i) {
    long long x1 = x - a[i].first.first, y1 = y - a[i].first.second;
    long long x2 = x + a[i].first.first, y2 = y + a[i].first.second;
    if (x1 * x1 + y1 * y1 < x2 * x2 + y2 * y2) {
      ans[a[i].second] ^= 1;
      x = x1, y = y1;
    } else {
      x = x2, y = y2;
    }
  }
  for (long long i = 0; i < (n); ++i) {
    cout << (ans[i] ? -1 : 1) << ' ';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  Solve();
  return 0;
}
