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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct Pt {
  long long x, y;
  Pt() {}
  Pt(long long x, long long y) : x(x), y(y) {}
  void read() { cin >> x >> y; }
  Pt operator+(Pt &p) { return Pt(x + p.x, y + p.y); }
  Pt operator-(Pt &p) { return Pt(x - p.x, y - p.y); }
  long long dist2() { return x * x + y * y; }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<Pt, int> > vec(n);
  for (int i = 0; i < int(n); ++i) {
    vec[i].first.read();
    vec[i].second = i;
  }
  vector<int> sign(n);
  Pt ans(10000000, 10000000);
  while (ans.dist2() > (long long)1500000 * 1500000) {
    ans = Pt(0, 0);
    for (int i = 0; i < n; ++i) {
      if ((ans + vec[i].first).dist2() < (ans - vec[i].first).dist2()) {
        sign[vec[i].second] = 1;
        ans = ans + vec[i].first;
      } else {
        sign[vec[i].second] = -1;
        ans = ans - vec[i].first;
      }
    }
    shuffle(vec.begin(), vec.end(), rng);
  }
  for (int i = 0; i < n; ++i) {
    cout << sign[i] << " ";
  }
  cout << '\n';
  return 0;
}
