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;
const int N = 2e5 + 10;
const double eps = 1e2, oo = 1.5e6;
struct Point {
  double x, y;
  double Len() { return sqrt(x * x + y * y); }
  Point operator+(const Point &rhs) const {
    return (Point){x + rhs.x, y + rhs.y, 0};
  }
  Point operator-(const Point &rhs) const {
    return (Point){x - rhs.x, y - rhs.y, 0};
  }
  int id;
} p[N];
int n, ans[N];
bool cmp(Point a, Point b) { return a.Len() > b.Len(); }
void dfs(int i, Point v) {
  if (i == n) {
    if (v.Len() < oo + eps) {
      for (int j = 0; j < n; j++) printf("%d ", ans[j]);
      exit(0);
    } else
      return;
  }
  Point pos = v + p[i], neg = v - p[i];
  if (pos.Len() - eps < neg.Len()) {
    ans[p[i].id] = 1;
    dfs(i + 1, pos);
  }
  if (neg.Len() - eps < pos.Len()) {
    ans[p[i].id] = -1;
    dfs(i + 1, neg);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lf%lf", &p[i].x, &p[i].y), p[i].id = i;
  sort(p, p + n, cmp);
  dfs(0, (Point){0, 0, 0});
  return 0;
}