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;
struct edge {
  long long a, b, cost;
  edge(long long x, long long y, long long z) : a(x), b(y), cost(z){};
};
struct data {
  long long x, y, id, ans;
};
bool comp(data a, data b) { return abs(a.x - a.y) > abs(b.x - b.y); }
bool comp1(data a, data b) { return a.id < b.id; }
long long dis(long long x, long long y) { return x * x + y * y; }
int main() {
  long long n;
  cin >> n;
  data a[n];
  for (long long i = 0; i < (long long)(n); i++) {
    a[i].id = i;
    cin >> a[i].x >> a[i].y;
  }
  long long p = 0, q = 0;
  sort(a, a + n, comp);
  for (long long i = (long long)(0); i <= (long long)(n - 1); i++) {
    if (dis(p + a[i].x, q + a[i].y) < dis(p - a[i].x, q - a[i].y))
      p += a[i].x, q += a[i].y, a[i].ans = 1;
    else
      p -= a[i].x, q -= a[i].y, a[i].ans = -1;
  }
  sort(a, a + n, comp1);
  for (long long i = (long long)(0); i <= (long long)(n - 1); i++)
    cout << a[i].ans << " ";
  return 0;
}