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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct Point {
  int x, y, index;
} points[100050];
int main() {
  int n, signs[100050];
  long long int xT = 0, yT = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    scanf("%d %d", &points[i].x, &points[i].y), points[i].index = i;
  while (true) {
    xT = 0;
    yT = 0;
    for (int i = 0; i < n; i++) {
      long long int xN = xT + points[i].x, yN = yT + points[i].y;
      long long int xN2 = xT - points[i].x, yN2 = yT - points[i].y;
      if (sqrt(xN * xN + yN * yN) <= sqrt(xN2 * xN2 + yN2 * yN2)) {
        signs[points[i].index] = 1;
        xT += points[i].x;
        yT += points[i].y;
      } else {
        signs[points[i].index] = -1;
        xT -= points[i].x;
        yT -= points[i].y;
      }
    }
    if (sqrt(xT * xT + yT * yT) <= 1.5e+6) break;
    random_shuffle(points, points + n);
  }
  for (int i = 0; i < n; i++) printf("%d ", signs[i]);
  return 0;
}