A11784 | Leaving the Bar

For a vector $\vec{v} = (x, y)$ , define $|v| = \sqrt{x^2 + y^2}$ .

Allen had a bit too much to drink at the bar, which is at the origin. There are $n$ vectors $\vec{v_1}, \vec{v_2}, \cdots, \vec{v_n}$ . 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 $|p| \le 1.5 \cdot 10^6$ so that he can stay safe.

The first line contains a single integer $n$ ( $1 \le n \le 10^5$ ) — 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 $|v_i| \le 10^6$ for all $i$ .

Output a single line containing $n$ integers $c_1, c_2, \cdots, c_n$ , each of which is either $1$ or $-1$ . Your solution is correct if the value of $p = \sum_{i = 1}^n c_i \vec{v_i}$ , satisfies $|p| \le 1.5 \cdot 10^6$ .

It can be shown that a solution always exists under the given constraints.