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;
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> b(n), xi(n), yi(n);
  for (int i = 0; i < n; ++i) {
    cin >> xi[i] >> yi[i];
    b[i] = i;
  }
  while (1) {
    random_shuffle(b.begin(), b.end());
    vector<int> ans(n);
    long long px = 0;
    long long py = 0;
    for (int i = 0; i < n; ++i) {
      int j = b[i];
      int x = xi[j], y = yi[j];
      long long u = (px + x) * (px + x) + (py + y) * (py + y);
      long long h = (px - x) * (px - x) + (py - y) * (py - y);
      if (u < h) {
        ans[j] = 1;
        px += x;
        py += y;
      } else {
        ans[j] = -1;
        px -= x;
        py -= y;
      }
    }
    if (px * px + py * py <= 1.5 * 1.5 * 1e12) {
      for (int i = 0; i < n; ++i) {
        cout << ans[i] << ' ';
      }
      break;
    }
  }
}