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;
int n;
struct node {
  int x, y, id;
  void scan(int id) {
    this->id = id;
    scanf("%d%d", &x, &y);
  }
} a[100010];
int ans[100010];
long long dis(long long x1, long long y1, long long x2, long long y2) {
  return (x1 + x2) * (x1 + x2) + (y1 + y2) * (y1 + y2);
}
int main() {
  srand(time(NULL));
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; ++i) a[i].scan(i);
    while (1) {
      random_shuffle(a + 1, a + 1 + n);
      long long x = 0, y = 0;
      for (int i = 1; i <= n; ++i) {
        if (dis(x, y, a[i].x, a[i].y) <= dis(x, y, -a[i].x, -a[i].y)) {
          x += a[i].x;
          y += a[i].y;
          ans[a[i].id] = 1;
        } else {
          x -= a[i].x;
          y -= a[i].y;
          ans[a[i].id] = -1;
        }
      }
      if (dis(0, 0, x, y) <= 225ll * 10000000000) break;
    }
    for (int i = 1; i <= n; ++i) printf("%d%c", ans[i], " \n"[i == n]);
  }
  return 0;
}