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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.StringTokenizer;

public class Code492 {
    public static final double EPS = 1e-9;
    public static class Pair {
        private double x;
        private double y;
        public Pair(double x, double y) {
            this.x = x;
            this.y = y;
        }
    }
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter w = new PrintWriter(System.out);
        int n = Integer.parseInt(br.readLine());
        long[] c = new long[n];
        Pair p = new Pair(0, 0);
        Pair[] pairs = new Pair[n];
        for (int i = 0; i < n; ++i) {
            String s = br.readLine();
            StringTokenizer st = new StringTokenizer(s);
            long x = Long.parseLong(st.nextToken());
            long y = Long.parseLong(st.nextToken());
            pairs[i] = new Pair(x, y);
            double xx = p.x + x;
            double yy = p.y + y;
            double pp = Math.sqrt(xx * xx + yy * yy);
            if (pp <= 1500000) {
                c[i] = 1;
                p.x += x;
                p.y += y;
            }else {
                c[i] = -1;
                p.x -= x;
                p.y -= y;
            }
        }
        while (Math.sqrt(p.x * p.x + p.y * p.y) >= 1500000) {
            for (int i = 0; i < n; ++i) {
                if (c[i] == -1) {
                    double xx = p.x + (pairs[i].x * 2);
                    double yy = p.y + (pairs[i].y * 2);
                    double ppp = Math.sqrt(p.x * p.x + p.y * p.y);
                    double pp = Math.sqrt(xx * xx + yy * yy);
                    if (pp <= ppp) {
                        c[i] = 1;
                        p.x += (pairs[i].x * 2);
                        p.y += (pairs[i].y * 2);
                        if (pp <= 1500000) {
                            break;
                        }
                    }
                }else {
                    double xx = p.x - (pairs[i].x * 2);
                    double yy = p.y - (pairs[i].y * 2);
                    double ppp = Math.sqrt(p.x * p.x + p.y * p.y);
                    double pp = Math.sqrt(xx * xx + yy * yy);
                    if (pp <= ppp) {
                        c[i] = -1;
                        p.x -= (pairs[i].x * 2);
                        p.y -= (pairs[i].y * 2);
                        if (pp <= 1500000) {
                            break;
                        }
                    }
                }
            }
        }
        for (int i = 0; i < n; ++i) {
            w.print(c[i] + " ");
        }
        w.close();
    }
}
