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 aux {
  int x, y, z;
} v[1000005];
int ans[1000005];
int main() {
  srand(time(NULL));
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int a, b;
    cin >> v[i].x >> v[i].y;
    v[i].z = i;
  }
  while (1) {
    int x = 0, y = 0;
    random_shuffle(v + 1, v + n + 1);
    for (int i = 1; i <= n; ++i) {
      int x1 = -1;
      int a = v[i].x;
      int b = v[i].y;
      if (1LL * (a + x) * (a + x) + 1LL * (b + y) * (b + y) >=
          1LL * (a - x) * (a - x) + 1LL * (b - y) * (b - y))
        x -= a, y -= b;
      else
        x += a, y += b, x1 = 1;
      ans[v[i].z] = x1;
    }
    if (1LL * x * x + 1LL * y * y <= 2250000000000LL) {
      for (int i = 1; i <= n; ++i) cout << ans[i] << ' ';
      return 0;
    }
  }
  return 0;
}