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;
long long gcd(long long a, long long b) { return (b ? gcd(b, a % b) : a); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long inf = 1e9 + 7;
long double first = 1500000;
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<pair<pair<long double, long long>, pair<long double, long double> > >
      v(n);
  for (long long i = 0; i < n; i++) {
    long double a, b;
    cin >> a >> b;
    v[i].first.first = sqrt(a * a + b * b);
    v[i].first.second = i;
    v[i].second.first = a;
    v[i].second.second = b;
  }
  unsigned seed = std::chrono::system_clock::now().time_since_epoch().count();
  shuffle(v.begin(), v.end(), std::default_random_engine(seed));
  vector<long long> ans(n, 0);
  long double x = 0, y = 0;
  for (long long i = 0; i < n; i++) {
    long double x1, x2, y1, y2;
    x1 = x + v[i].second.first;
    x2 = x - v[i].second.first;
    y1 = y + v[i].second.second;
    y2 = y - v[i].second.second;
    if (x1 * x1 + y1 * y1 > x2 * x2 + y2 * y2) {
      if (x1 * x1 + y1 * y1 <= first) {
        x = x1;
        y = y1;
        ans[v[i].first.second] = 1;
      } else {
        x = x2;
        y = y2;
        ans[v[i].first.second] = -1;
      }
    } else {
      if (x2 * x2 + y2 * y2 <= first) {
        x = x2;
        y = y2;
        ans[v[i].first.second] = -1;
      } else {
        x = x1;
        y = y1;
        ans[v[i].first.second] = 1;
      }
    }
  }
  for (long long i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
}