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 binexp(long long a, long long b, long long m) {
  long long res = 1;
  a = a % m;
  while (b) {
    if (b & 1) res = (a * 1LL * res) % m;
    a = (a * 1LL * a) % m;
    b >>= 1;
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  pair<pair<long long, long long>, long long> p[n];
  for (long long i = 0; i < n; i++) {
    cin >> p[i].first.first >> p[i].first.second;
    p[i].second = i;
  }
  while (1) {
    random_shuffle(p, p + n);
    long long px = 0, py = 0;
    long long answ[n];
    for (long long i = 0; i < n; i++) {
      long long u1 = px + p[i].first.first, v1 = py + p[i].first.second,
                u2 = px - p[i].first.first, v2 = py - p[i].first.second;
      if (u1 * u1 + v1 * v1 <= u2 * u2 + v2 * v2) {
        answ[p[i].second] = 1;
        px = u1, py = v1;
      } else {
        answ[p[i].second] = -1;
        px = u2, py = v2;
      }
    }
    if (px * px + py * py <= 2250000000000LL) {
      for (long long i = 0; i < n; i++) cout << answ[i] << " ";
      return 0;
    }
  }
  return 0;
}