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;
struct position {
  unsigned long long first;
  unsigned long long second;
  int id;
  position() {}
  position(int a, int b) {
    first = a;
    second = b;
  }
};
int main() {
  srand(time(NULL));
  int n;
  cin >> n;
  vector<position> vec(n);
  for (int i = 0; i < n; ++i) {
    cin >> vec[i].first >> vec[i].second;
    vec[i].id = i;
  }
  while (true) {
    vector<int> res(n);
    position pos(0, 0);
    random_shuffle(vec.begin(), vec.end());
    for (int i = 0; i < n; i++) {
      unsigned long long d1 =
          (vec[i].first + pos.first) * (vec[i].first + pos.first) +
          (vec[i].second + pos.second) * (vec[i].second + pos.second);
      unsigned long long d2 =
          (vec[i].first - pos.first) * (vec[i].first - pos.first) +
          (vec[i].second - pos.second) * (vec[i].second - pos.second);
      if (d1 < d2) {
        pos.first += vec[i].first;
        pos.second += vec[i].second;
        res[vec[i].id] = 1;
      } else {
        pos.first -= vec[i].first;
        pos.second -= vec[i].second;
        res[vec[i].id] = -1;
      }
    }
    if (sqrt(pos.first * pos.first + pos.second * pos.second) < 1500000) {
      for (int i = 0; i < n; ++i) {
        cout << res[i] << " ";
      }
      cout << endl;
      break;
    }
  }
  return 0;
}