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.Random;
import java.util.Scanner;

public class e implements Runnable{
	public static void main(String[] args) {
		new Thread(null, new e(), ":9", 1<<27).start();
	}
	public void run(){
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();
		long[][] arr = new long[N][];
		for(int n=0;n<N;n++){
//			arr[n] = new long[]{	
//			(long)(Math.random() * 2_000_001) - 1_000_000,
//			(long)(Math.random() * 2_000_001) - 1_000_000};
			arr[n] = new long[]{in.nextLong(), in.nextLong()};
		}
		
		// first let's try going the farthest away
		long[] sum = new long[2];
		boolean[] flipped = new boolean[N];
		for(int n=0;n<N;n++){
			if(arr[n][0] < 0 || (arr[n][0] == 0 && arr[n][1] < 0)){
				
				flipped[n] = !flipped[n];
				arr[n][0] = -arr[n][0];
				arr[n][1] = -arr[n][1];
			}

			sum[0] += arr[n][0];
			sum[1] += arr[n][1];
		}
		
		// let's do the following until we're right
		// EDIT: okay now it should be difficult for people to come up with precision errors.
		// take that, hackers.
		for(;;){
			// now what?
			// well the direction of the sum is the OPPOSITE of where we want to go
			// so umm, let's inch towards that place
			// and use dot product to check if adding that makes us closer
			for(int n=0;n<N;n++){
				
				double chance;
				if(arr[n][0]*sum[0]+arr[n][1]*sum[1] <= 0){
					// i thought having a negative dot product makes the probability higher
					// but empirical tests suggest otherwise
					chance = 0.1;
				}
				else{
					chance = 0.9;
				}
				
				if(Math.random() < chance){
					sum[0] -= 2*arr[n][0];
					sum[1] -= 2*arr[n][1];
					
					arr[n][0] = -arr[n][0];
					arr[n][1] = -arr[n][1];
					flipped[n] = !flipped[n];
				}
				
				if(withinDist(sum)) break;
			}
			if(withinDist(sum)) break;
//			System.err.println(Math.hypot(sum[0], sum[1]));
//			System.err.println("ayy");
		}

//		System.err.println(Math.hypot(sum[0], sum[1]));
		StringBuilder sb = new StringBuilder();
		for(int n=0;n<N;n++){
			if(n > 0) sb.append(' ');
			sb.append(flipped[n]? "1": "-1");
		}
		System.out.println(sb);
	}
	
	
	long DIST = 1_500_000L;
	boolean withinDist(long[] sum){
		return Math.hypot(sum[0], sum[1]) < DIST;
	}
}