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 P {
  int x, y, id;
  P() {}
  P(int x, int y, int id) : x(x), y(y), id(id) {}
} p[100100];
long long cal(int x, int y) { return 1LL * x * x + 1LL * y * y; }
int b[100100];
int main() {
  ios::sync_with_stdio(0);
  srand(time(0));
  int n, m;
  while (cin >> n) {
    for (int i = 0; i < n; i++) cin >> p[i].x >> p[i].y, p[i].id = i;
    random_shuffle(p, p + n);
    P ans(0, 0, 0);
    for (int i = 0; i < n; i++) {
      if (cal(ans.x + p[i].x, ans.y + p[i].y) <
          cal(ans.x - p[i].x, ans.y - p[i].y)) {
        b[p[i].id] = 1;
        ans.x += p[i].x;
        ans.y += p[i].y;
      } else {
        b[p[i].id] = -1;
        ans.x -= p[i].x;
        ans.y -= p[i].y;
      }
    }
    for (int i = 0; i < n; i++) printf("%d%c", b[i], " \n"[i == n - 1]);
  }
  return 0;
}