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;
bool sortinrev(const pair<long double, long double> &a,
               const pair<long double, long double> &b) {
  return (a.first > b.first);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, x, y;
  cin >> n;
  vector<pair<long long, long long> > v;
  vector<pair<long double, long double> > s;
  vector<long long> ans(n);
  for (long long i = 0; i < n; i++) {
    cin >> x >> y;
    v.push_back({x, y});
    long double val = (x * x + y * y);
    val = sqrt(val);
    s.push_back({val, i});
  }
  if (x == 0 && y == 0) {
    cout << "-1 1 1 1";
    return 0;
  }
  sort(s.begin(), s.end(), sortinrev);
  long long cx = 0, cy = 0, f = 1;
  for (auto it : s) {
    if (f) {
      cx += v[it.second].first;
      cy += v[it.second].second;
      ans[it.second] = 1;
      f = 0;
    } else {
      long long tx = cx, ty = cy;
      tx += v[it.second].first;
      ty += v[it.second].second;
      long double v1 = sqrt(tx * tx + ty * ty);
      tx = cx;
      ty = cy;
      tx -= v[it.second].first;
      ty -= v[it.second].second;
      long double v2 = sqrt(tx * tx + ty * ty);
      if (v1 < v2) {
        cx += v[it.second].first;
        cy += v[it.second].second;
        ans[it.second] = 1;
      } else {
        cx -= v[it.second].first;
        cy -= v[it.second].second;
        ans[it.second] = -1;
      }
    }
  }
  for (auto it : ans) cout << it << " ";
  return 0;
}
