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;
signed main() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  vector<pair<pair<long long, long long>, long long>> res(n);
  for (long long i = 0; i < n; i++) {
    cin >> res[i].first.first >> res[i].first.second;
    res[i].second = i;
    if (res[i].first.first < 0) {
      res[i].first.first *= -1;
      res[i].first.second *= -1;
      a[i] ^= 1;
    }
  }
  sort(res.begin(), res.end(),
       [&](pair<pair<long long, long long>, long long>& aa,
           pair<pair<long long, long long>, long long>& bb) {
         return aa.first.second * bb.first.first <
                aa.first.first * bb.first.second;
       });
  long long x = res[0].first.first;
  long long y = res[0].first.second;
  for (long long i = 1; i < n; i++) {
    long long u = x - res[i].first.first;
    long long d = y - res[i].first.second;
    long long l = x + res[i].first.first;
    long long r = y + res[i].first.second;
    if (u * u + d * d < l * l + r * r) {
      a[res[i].second] ^= 1;
      x = u;
      y = d;
    } else {
      x = l;
      y = r;
    }
  }
  for (long long i = 0; i < a.size(); i++) cout << (a[i] ? -1 : 1) << ' ';
  return 0;
}
