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 

import java.awt.Point;
import java.util.Scanner;
import java.util.SortedSet;

public class E {

    static boolean merre(Point p1, Point p2) {
        return Point.distance(p1.x + p2.x, p1.y + p2.y, 0, 0)
                < Point.distance(p1.x - p2.x, p1.y - p2.y, 0, 0);
    }

    static int max = 1500000;

    public static void main(String args[]) {
        Scanner s = new Scanner(System.in);
        int N = s.nextInt();
        Point p[] = new Point[N];
        int ir[] = new int[N];
        Point H = new Point(0, 0);
        for (int i = 0; i < N; i++) {
            p[i] = new Point(s.nextInt(), s.nextInt());
            if (merre(H, p[i])) {
                ir[i] = 1;
                H.x += p[i].x;
                H.y += p[i].y;
            } else {
                ir[i] = -1;
                H.x -= p[i].x;
                H.y -= p[i].y;
            }
        }
        while (Point.distance(H.x, H.y, 0, 0) > (max)) {
            for (int i = 0; i < N; i++) {
                if (Point.distance(H.x, H.y, 0, 0)
                        > Point.distance(H.x - 2 * ir[i] * p[i].x, H.y - 2 * ir[i] * p[i].y, 0, 0)) {
                    H.x -= 2 * ir[i] * p[i].x;
                    H.y -= 2 * ir[i] * p[i].y;
                    ir[i] *= -1;
                }
            }
        }
           for (int i = 0; i < N; i++) {
               System.out.print(ir[i]+" ");
           }
    }
}
