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;
const int N = 1e6 + 5;
bool comp(pair<pair<long long, long long>, int> X,
          pair<pair<long long, long long>, int> Y) {
  long long dis1 = abs(X.first.second * Y.first.first),
            dis2 = abs(Y.first.second * X.first.first);
  if (dis1 == dis2)
    return X.second < Y.second;
  else
    return dis1 < dis2;
}
bool isLess(pair<long long, long long> a, pair<long long, long long> b) {
  return a.first * a.first + a.second * a.second <
         b.first * b.first + b.second * b.second;
}
int ans[N];
vector<pair<pair<long long, long long>, int> > D;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  pair<long long, long long> at = make_pair(0, 0);
  for (int i = 1; i <= n; i++) {
    int x, y;
    cin >> x >> y;
    D.push_back(make_pair(make_pair(x, y), i));
  }
  sort(D.begin(), D.end(), comp);
  for (auto it : D) {
    pair<long long, long long> at1 =
        make_pair(at.first + it.first.first, at.second + it.first.second);
    pair<long long, long long> at2 =
        make_pair(at.first - it.first.first, at.second - it.first.second);
    if (isLess(at1, at2)) {
      ans[it.second] = 1;
      at = at1;
    } else {
      ans[it.second] = -1;
      at = at2;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
  return 0;
}
