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;
const int N = 2e5 + 5;
const long long int BOUND = 1.5e6;
const long long int SQ_BOUND = BOUND * BOUND;
vector<complex<long long int> > arr;
long long int s[N];
inline long long int norm_sq(const complex<long long int> &z) {
  return (z * conj(z)).real();
}
long long int greedy(long long int sgn) {
  complex<long long int> o{0, 0};
  reverse((arr).begin(), (arr).end());
  o = (s[0] = sgn) * arr[0];
  for (int i = 1; i < (int)arr.size(); ++i) {
    complex<long long int> po = o + arr[i];
    complex<long long int> ne = o - arr[i];
    if (norm_sq(po) <= norm_sq(ne)) {
      s[i] = 1;
      o = po;
    } else {
      s[i] = -1;
      o = ne;
    }
  }
  reverse((arr).begin(), (arr).end());
  return norm_sq(o);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  srand(time(NULL));
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    arr.emplace_back(x, y);
  }
  long long int cur, sgn = 1;
  do {
    cur = greedy(sgn);
    if (!~sgn) random_shuffle((arr).begin(), (arr).end());
    sgn = -sgn;
  } while (cur > SQ_BOUND);
  for (int i = n - 1; i >= 0; --i) cout << s[i] << " \n"[i + 1 == n];
  return 0;
}