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 {
  int index;
  long long x, y;
  Point(long long a = 0, long long b = 0, int index = 0) {
    x = a, y = b, index = 0;
  }
  Point operator+(const Point &P) { return Point(x + P.x, y + P.y); }
  Point operator-(const Point &P) { return Point(x - P.x, y - P.y); }
};
long long square(long long n) { return n * n; }
long long square_norm(Point P) { return (P.x * P.x + P.y * P.y); }
int main() {
  int n;
  scanf("%d", &n);
  vector<Point> V(n);
  for (int i = 0; i < n; i++) {
    scanf("%I64d %I64d", &V[i].x, &V[i].y);
    V[i].index = i;
  }
  const long long oo = 1.5e6;
  Point P;
  vector<int> step(n);
  do {
    random_shuffle((V).begin(), (V).end());
    P = Point(0, 0);
    for (int i = 0; i < n; i++) {
      if (square_norm(P + V[i]) <= square_norm(P - V[i])) {
        P = P + V[i];
        step[V[i].index] = 1;
      } else {
        P = P - V[i];
        step[V[i].index] = -1;
      }
    }
  } while (square_norm(P) > square(oo));
  for (int i = 0; i < n; i++) printf("%d ", step[i]);
  return 0;
}