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);
  int n, i;
  cin >> n;
  int ans[n];
  long long int x, y, d, d1, d2, d3, d4;
  vector<pair<long long int, long long int> > v, p;
  pair<long long int, long long int> c = make_pair(0, 0);
  for (i = 0; i < n; i++) {
    cin >> x >> y;
    d = sqrt(pow(x, 2) + pow(y, 2));
    v.emplace_back(make_pair(d, i));
    p.emplace_back(make_pair(x, y));
  }
  sort((v).begin(), (v).end());
  c.first += p[v[n - 1].second].first;
  c.second += p[v[n - 1].second].second;
  ans[v[n - 1].second] = 1;
  for (i = n - 2; i >= 0; i--) {
    d1 = sqrt(pow(c.first + p[v[i].second].first, 2) +
              pow(c.second + p[v[i].second].second, 2));
    d2 = sqrt(pow(c.first - p[v[i].second].first, 2) +
              pow(c.second - p[v[i].second].second, 2));
    if (i == 0) {
      if (d1 < d2) {
        ans[v[i].second] = 1;
        c.first = c.first + p[v[i].second].first;
        c.second = c.second + p[v[i].second].second;
      } else {
        c.first = c.first - p[v[i].second].first;
        c.second = c.second - p[v[i].second].second;
        ans[v[i].second] = -1;
      }
    } else {
      d3 = sqrt(pow(c.first + p[v[i - 1].second].first, 2) +
                pow(c.second + p[v[i - 1].second].second, 2));
      d4 = sqrt(pow(c.first - p[v[i - 1].second].first, 2) +
                pow(c.second - p[v[i - 1].second].second, 2));
      if (d1 <= d2 && d1 <= d3 && d1 <= d4) {
        ans[v[i].second] = 1;
        c.first = c.first + p[v[i].second].first;
        c.second = c.second + p[v[i].second].second;
      } else if (d2 <= d1 && d2 <= d3 && d2 <= d4) {
        ans[v[i].second] = -1;
        c.first = c.first - p[v[i].second].first;
        c.second = c.second - p[v[i].second].second;
      } else if (d3 <= d4 && d3 <= d2 && d3 <= d1) {
        v[i - 1].swap(v[i]);
        ans[v[i].second] = 1;
        c.first = c.first + p[v[i].second].first;
        c.second = c.second + p[v[i].second].second;
      } else {
        v[i - 1].swap(v[i]);
        ans[v[i].second] = -1;
        c.first = c.first - p[v[i].second].first;
        c.second = c.second - p[v[i].second].second;
      }
    }
  }
  for (i = 0; i < n; i++) cout << ans[i] << " ";
  return 0;
}
