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;
const int N = 100050;
const long long goal = (long long)1500000 * 1500000;
struct Vector {
  int x, y;
  Vector() {}
  long long sz() { return (long long)x * x + (long long)y * y; }
  Vector operator+(Vector b) const {
    Vector c;
    c.x = x + b.x;
    c.y = y + b.y;
    return c;
  }
  Vector operator-(Vector b) const {
    Vector c;
    c.x = x - b.x;
    c.y = y - b.y;
    return c;
  }
} a[N];
bool operator<(Vector a, Vector b) { return a.sz() <= b.sz(); }
int id[N], sol[N];
int main() {
  srand(time(0));
  int n, i;
  scanf("%i", &n);
  for (i = 1; i <= n; i++) scanf("%i %i", &a[i].x, &a[i].y);
  for (i = 1; i <= n; i++) id[i] = i;
  while (1) {
    random_shuffle(id + 1, id + 1 + n);
    Vector my;
    my.x = my.y = 0;
    for (i = 1; i <= n; i++) {
      if ((my + a[id[i]]) < (my - a[id[i]])) {
        my = my + a[id[i]];
        sol[id[i]] = 1;
      } else
        my = my - a[id[i]], sol[id[i]] = -1;
    }
    if (my.sz() <= goal) break;
  }
  for (i = 1; i <= n; i++) printf("%i ", sol[i]);
  return 0;
}