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 ans[212345];
vector<pair<pair<long long, long long>, long long> > vec;
bool cmp(pair<pair<long long, long long>, long long> a,
         pair<pair<long long, long long>, long long> b) {
  return pow(a.first.first, 2) + pow(a.first.second, 2) >
         pow(b.first.first, 2) + pow(b.first.second, 2);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    vec.push_back(make_pair(make_pair(a, b), i));
  }
  while (1) {
    long long num = 0, x = 0, y = 0;
    for (int i = 0; i < n; i++) {
      if ((x + vec[i].first.first) * (x + vec[i].first.first) +
              (y + vec[i].first.second) * (y + vec[i].first.second) >
          (x - vec[i].first.first) * (x - vec[i].first.first) +
              (y - vec[i].first.second) * (y - vec[i].first.second)) {
        x -= vec[i].first.first;
        y -= vec[i].first.second;
        ans[vec[i].second] = -1;
      } else {
        x += vec[i].first.first;
        y += vec[i].first.second;
        ans[vec[i].second] = 1;
      }
    }
    if (x * x + y * y <= 225e10) {
      break;
    } else {
      random_shuffle(vec.begin(), vec.end());
    }
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] << ' ';
  }
  return 0;
}