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() {
  long long n;
  cin >> n;
  vector<pair<long long, long long> > v(n), vs(n);
  int arr[n];
  memset(arr, 0, sizeof arr);
  for (long long i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second;
    if (i == 0) {
      arr[i] = 1;
      vs[i].first = v[i].first;
      vs[i].second = v[i].second;
    }
    if (i > 0) {
      long long valx = vs[i - 1].first;
      long long valy = vs[i - 1].second;
      long long v1 = (((valx + v[i].first) * (valx + v[i].first)) +
                      ((valy + v[i].second) * (valy + v[i].second)));
      long long v2 = (((valx - v[i].first) * (valx - v[i].first)) +
                      ((valy - v[i].second) * (valy - v[i].second)));
      if (v1 <= v2) {
        arr[i] = 1;
        vs[i].first = (valx + v[i].first);
        vs[i].second = (valy + v[i].second);
      } else {
        arr[i] = -1;
        vs[i].first = (valx - v[i].first);
        vs[i].second = (valy - v[i].second);
      }
    }
  }
  if ((vs[n - 1].first * vs[n - 1].first) +
          (vs[n - 1].second * vs[n - 1].second) >
      2250000000000) {
    long long i = 0;
    while ((vs[n - 1].first * vs[n - 1].first) +
               (vs[n - 1].second * vs[n - 1].second) >
           2250000000000) {
      if (i > n) {
        i = i % n;
      }
      long long valx = vs[n - 1].first - ((2 * arr[i]) * v[i].first);
      long long valy = vs[n - 1].second - ((2 * arr[i]) * v[i].second);
      if ((vs[n - 1].first * vs[n - 1].first) +
              (vs[n - 1].second * vs[n - 1].second) >
          ((valx * valx) + (valy * valy))) {
        arr[i] = -arr[i];
        vs[n - 1].first = valx;
        vs[n - 1].second = valy;
      }
      i++;
    }
  }
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  return 0;
}