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 point {
  long long x, y, idx;
} p[100007], base;
int sign(point p, point q, point r) {
  long long val = (q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y);
  val = -val;
  if (val == 0) return 0;
  if (val < 0) return 1;
  if (val > 0) return 2;
}
long long val(point a) { return a.x * a.x + a.y * a.y; }
int main() {
  base.x = 0;
  base.y = 0;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    scanf("%lld %lld", &p[i].x, &p[i].y), p[i].idx = i;
  vector<int> ans(n);
  int cnt = 0;
  do {
    cnt++;
    random_shuffle(p, p + n);
    base.x = 0;
    base.y = 0;
    for (int i = 0; i < n; i++) {
      point P, Q;
      P.x = p[i].x + base.x;
      P.y = p[i].y + base.y;
      Q.x = base.x - p[i].x;
      Q.y = base.x - p[i].y;
      if (val(P) <= val(Q)) {
        ans[p[i].idx] = 1;
        base.x += p[i].x;
        base.y += p[i].y;
      } else {
        ans[p[i].idx] = -1;
        base.x -= p[i].x;
        base.y -= p[i].y;
      }
    }
    if (val(base) <= 1500000LL * 1500000LL)
      for (int i = 0; i < n; i++) printf("%d ", ans[i]);
  } while (val(base) > 1500000LL * 1500000LL);
}