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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  long long x[n], y[n];
  long long sx, sy;
  int sign[n];
  double val, comp1, comp2;
  for (int i = 0; i < n; ++i) cin >> x[i] >> y[i];
  vector<pair<double, int> > v;
  for (int i = 0; i < n; ++i) {
    val = sqrt(x[i] * x[i] + y[i] * y[i]);
    v.push_back({val, i});
  }
  sort(v.begin(), v.end());
  sign[v[n - 1].second] = 1;
  sx = x[v[n - 1].second], sy = y[v[n - 1].second];
  for (int i = n - 2; i >= 0; --i) {
    comp1 = sqrt((sx + x[v[i].second]) * (sx + x[v[i].second]) +
                 (sy + y[v[i].second]) * (sy + y[v[i].second]));
    comp2 = sqrt((sx - x[v[i].second]) * (sx - x[v[i].second]) +
                 (sy - y[v[i].second]) * (sy - y[v[i].second]));
    if (comp1 > comp2) {
      sign[v[i].second] = -1;
      sx = sx - x[v[i].second];
      sy = sy - y[v[i].second];
    } else {
      sign[v[i].second] = 1;
      sx = sx + x[v[i].second];
      sy = sy + y[v[i].second];
    }
  }
  if (sqrt(sx * sx + sy * sy) > 1500000) {
    sign[v[0].second] = 1;
    sx = x[v[0].second], sy = y[v[0].second];
    for (int i = 1; i < n; ++i) {
      comp1 = sqrt((sx + x[v[i].second]) * (sx + x[v[i].second]) +
                   (sy + y[v[i].second]) * (sy + y[v[i].second]));
      comp2 = sqrt((sx - x[v[i].second]) * (sx - x[v[i].second]) +
                   (sy - y[v[i].second]) * (sy - y[v[i].second]));
      if (comp1 > comp2) {
        sign[v[i].second] = -1;
        sx = sx - x[v[i].second];
        sy = sy - y[v[i].second];
      } else {
        sign[v[i].second] = 1;
        sx = sx + x[v[i].second];
        sy = sy + y[v[i].second];
      }
    }
  }
  for (int i = 0; i < n; ++i) cout << sign[i] << ' ';
}
