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;
constexpr long long N = 2e5 + 5;
constexpr long long M = 998244353;
long long n;
pair<long long, long long> a[N];
long long pos[N];
long long ans[N];
long long len(pair<long long, long long> p) {
  return p.first * p.first + p.second * p.second;
}
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  cin >> n;
  for (long long i = 1; i <= n; i++)
    cin >> a[i].first >> a[i].second, pos[i] = i, ans[i] = 1;
  pair<long long, long long> p = {0, 0};
  for (long long i = n; i >= 1; i--) {
    if (len({p.first + a[i].first, p.second + a[i].second}) <
        len({p.first - a[i].first, p.second - a[i].second})) {
      ans[i] = 1;
      p.first += a[i].first, p.second += a[i].second;
    } else {
      ans[i] = -1;
      p.first -= a[i].first, p.second -= a[i].second;
    }
  }
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
