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.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class Main {

	static BufferedReader in;
	static PrintWriter out;
	static StringTokenizer tok;

	class Point {
		public Point(int x, int y) {
			this.x = x;
			this.y = y;
		}

		public Point(int op, Point p1, Point p2) {
			this.op = op;
			this.p1 = p1;
			this.p2 = p2;
			if (op == 0) {
				x = p1.x + p2.x;
				y = p1.y + p2.y;
			} else {
				x = p1.x - p2.x;
				y = p1.y - p2.y;
			}
		}

		int x, y;
		Point p1, p2;
		int op;

		void dfs(int prev) {
			if (p1 != null)
				p1.dfs(prev);
			if (p2 != null)
				p2.dfs(op == 1 ? -prev : prev);
			op = prev;
		}

		public String toString() {
			return "[" + x + "," + y + "]";
		}
	};

	void solve() throws IOException {
		solve(ni());
	}

	int f[];
	List<Point> l = new ArrayList<Point>();

	private void solve(int n) throws IOException {
		if (n == 1) {
			out.println("1");
			return;
		}
		for (int i = 0; i < n; i++)
			l.add(new Point(ni(), ni()));
		List<Point> l2 = new ArrayList<Point>(l);

		Point[] pts = null;
		if (l2.size() > 2) {
			while (!l2.isEmpty()) {
				if (pts == null)
					pts = merge(l2.remove(l2.size() - 1), l2.remove(l2.size() - 1), l2.remove(l2.size() - 1));
				else
					pts = merge(l2.remove(l2.size() - 1), pts[0], pts[1]);
			}
		} else
			pts = l2.toArray(new Point[0]);
		merge2(pts[0], pts[1]).dfs(1);
		for (Point p : l)
			out.print(p.op + " ");
	}

	private Point[] merge(Point p1, Point p2, Point p3) {
//		Helper.tr(p1, p2, p3);
		Point p0 = merge(p1, p2);
		if (p0 != null)
			return new Point[] { p0, p3 };
		p0 = merge(p3, p2);
		if (p0 != null)
			return new Point[] { p0, p1 };
		p0 = merge(p3, p1);
		if (p0 != null)
			return new Point[] { p0, p2 };
		throw new RuntimeException("");
	}

	private Point merge(Point p1, Point p2) {
//		Helper.tr(p1, p2);
		long l1 = ((long) (p1.x + p2.x) * (p1.x + p2.x) + (long) (p1.y + p2.y) * (p1.y + p2.y));
		if (l1 <= 1000000000000L)
			return new Point(0, p1, p2);
//		Helper.tr((long)Math.sqrt(l1));
		l1 = ((long) (p1.x - p2.x) * (p1.x - p2.x) + (long) (p1.y - p2.y) * (p1.y - p2.y));
		if (l1 <= 1000000000000L)
			return new Point(1, p1, p2);
//		Helper.tr((long)Math.sqrt(l1));
		return null;
	}

	private Point merge2(Point p1, Point p2) {
		long l1 = ((long) (p1.x + p2.x) * (p1.x + p2.x) + (long) (p1.y + p2.y) * (p1.y + p2.y));
		long l2 = ((long) (p1.x - p2.x) * (p1.x - p2.x) + (long) (p1.y - p2.y) * (p1.y - p2.y));
		if (l1 < l2)
			return new Point(0, p1, p2);
		return new Point(1, p1, p2);
	}

	String ns() throws IOException {
		while (!tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine(), " ");
		}
		return tok.nextToken();
	}

	int ni() throws IOException {
		return Integer.parseInt(ns());
	}

	long nl() throws IOException {
		return Long.parseLong(ns());
	}

	double nd() throws IOException {
		return Double.parseDouble(ns());
	}

	String[] nsa(int n) throws IOException {
		String[] res = new String[n];
		for (int i = 0; i < n; i++) {
			res[i] = ns();
		}
		return res;
	}

	int[] nia(int n) throws IOException {
		int[] res = new int[n];
		for (int i = 0; i < n; i++) {
			res[i] = ni();
		}
		return res;
	}

	long[] nla(int n) throws IOException {
		long[] res = new long[n];
		for (int i = 0; i < n; i++) {
			res[i] = nl();
		}
		return res;
	}

	public static void main(String[] args) throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		tok = new StringTokenizer("");
		Main main = new Main();
		main.solve();
		out.close();
	}
}