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;
int main() {
  int n;
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  vector<tuple<long long, long long, int> > v(n);
  vector<int> ans(n);
  for (int i = 0; i < n; ++i) {
    long long a, b;
    cin >> a >> b;
    v[i] = make_tuple(a, b, i);
  }
  long long safe = 1500000ll * 1500000ll;
  while (1) {
    random_shuffle((v).begin(), (v).end());
    long long x = 0, y = 0;
    for (int(i) = (0); (i) < (n); ++(i)) {
      long long vx = get<0>(v[i]), vy = get<1>(v[i]);
      if ((x + vx) * (x + vx) + (y + vy) * (y + vy) >
          (x - vx) * (x - vx) + (y - vy) * (y - vy)) {
        ans[get<2>(v[i])] = -1;
        x -= vx;
        y -= vy;
      } else {
        ans[get<2>(v[i])] = 1;
        x += vx;
        y += vy;
      }
    }
    if (x * x + y * y < safe) {
      for (const auto& k : ans) cout << k << " ";
      cout << endl;
      return 0;
    }
  }
}
