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;
const int MAX = 5e5 + 3;
const long long MOD = 998244353;
const long long LIM = 1500000ll * 1500000ll;
pair<int, int> operator+(const pair<int, int> &a, const pair<int, int> &b) {
  return pair<int, int>(a.first + b.first, a.second + b.second);
}
pair<int, int> operator-(const pair<int, int> &a, const pair<int, int> &b) {
  return pair<int, int>(a.first - b.first, a.second - b.second);
}
long long norm2(pair<int, int> pos) {
  return 1ll * pos.first * pos.first + 1ll * pos.second * pos.second;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  srand(time(0));
  int n;
  cin >> n;
  vector<pair<int, int> > v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second;
  }
  vector<int> order(n);
  vector<int> ans(n);
  iota(order.begin(), order.end(), 0);
  pair<int, int> pos;
  do {
    pos = pair<int, int>(0, 0);
    random_shuffle(order.begin(), order.end());
    for (int i = 0; i < n; i++) {
      int t = order[i];
      if (norm2(pos + v[t]) < norm2(pos - v[t])) {
        pos = pos + v[t];
        ans[t] = 1;
      } else {
        pos = pos - v[t];
        ans[t] = -1;
      }
    }
  } while (norm2(pos) > LIM);
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
}