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 double esp = 1e-6;
const int MAXN = 1e5 + 5;
const long long LIMIT = 2250000000000;
struct Vector {
  long long x, y;
  int ind;
  Vector(long long _x = 0, long long _y = 0) { x = _x, y = _y; }
  void Input(int ti) {
    scanf("%lld %lld", &x, &y);
    ind = ti;
  }
  void Output() { printf("%lld %lld\n", x, y); }
  long long distance() { return x * x + y * y; }
};
Vector operator+(Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator-(Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
int ans[MAXN];
Vector inp[MAXN];
int main() {
  int n;
  scanf("%d", &n);
  Vector now;
  Vector t1, t2;
  for (int i = 0; i < n; i++) inp[i].Input(i);
  while (1) {
    now = Vector(0, 0);
    for (int i = 0; i < n; i++) {
      t1 = now + inp[i];
      t2 = now - inp[i];
      if (t1.distance() < t2.distance()) {
        ans[inp[i].ind] = 1;
        now = t1;
      } else {
        ans[inp[i].ind] = -1;
        now = t2;
      }
    }
    if (now.distance() <= LIMIT) break;
    random_shuffle(inp, inp + n);
  }
  for (int i = 0; i < n; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}