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 JAVA):

Solution:

import java.util.*;

public class hoyejabe
{
    static Scanner io = new Scanner(System.in);
    static int n, ans[];
    static Integer arr[];
    static Vector[] vectors;

    public static void main(String[] args)
    {
        int i;
        vectors = new Vector[n = io.nextInt()];
        ans = new int[n];
        arr = new Integer[n];

        for(i = 0; i < n; ++i) {
            vectors[i] = new Vector(io.nextDouble(), io.nextDouble());
            arr[i] = i;
        }

        Arrays.sort(arr, (Integer a, Integer b) -> -Double.compare(vectors[a].m, vectors[b].m));

        Vector p = new Vector(0, 0);
        for(i = 0; i < n; ++i)
        {
            Vector add = p.add(vectors[arr[i]]), sub = p.sub(vectors[arr[i]]);
            if(add.m > sub.m)
            {
                ans[arr[i]] = -1;
                p = sub;
            }
            else
            {
                ans[arr[i]] = 1;
                p = add;
            }
        }
        while(p.m > 225e10)
        {
            Vector temp;
            for(i = 0; i < n; ++i)
            {
                if(ans[i] > 0)
                {
                    temp = p.sub(vectors[i]).sub(vectors[i]);
                    if(temp.m < p.m)
                    {
                        p = temp;
                        ans[i] = -1;
                    }
                }
                else
                {
                    temp = p.add(vectors[i]).add(vectors[i]);
                    if(temp.m < p.m)
                    {
                        p = temp;
                        ans[i] = 1;
                    }
                }
            }
        }
        for(int jk: ans)
            System.out.print(jk + " ");
    }
}

class Vector
{
    double x, y, m;
    Vector(double a, double b)
    {
        x = a;  y = b;
        m = x*x + y*y;
    }
    Vector add(Vector v)
    {
        return new Vector(x + v.x, y + v.y);
    }
    Vector sub(Vector v)
    {
        return new Vector(x - v.x, y - v.y);
    }
}