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 
#include <bits/stdc++.h>
using namespace std;
struct kdd {
  int xx, yy, d;
};
bool ss(const pair<pair<long long, long long>, int> &u,
        const pair<pair<long long, long long>, int> &v) {
  return u.first.second * v.first.first < u.first.first * v.first.second;
}
int n;
pair<pair<long long, long long>, int> e[100001];
short c[100001];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d%I64d", &e[i].first.first, &e[i].first.second);
    if (e[i].first.first >= 0)
      c[i] = 1;
    else
      c[i] = -1, e[i].first.first = -e[i].first.first,
      e[i].first.second = -e[i].first.second;
    e[i].second = i;
  }
  sort(e + 1, e + 1 + n, ss);
  int u = e[1].first.first, v = e[1].first.second;
  for (int i = 2; i <= n; ++i) {
    long long uk = e[i].first.first, vk = e[i].first.second;
    if ((u + uk) * (u + uk) + (v + vk) * (v + vk) >
        (u - uk) * (u - uk) + (v - vk) * (v - vk))
      u -= uk, v -= vk, c[e[i].second] = -c[e[i].second];
    else
      u += uk, v += vk;
  }
  for (int i = 1; i <= n; ++i) printf("%hd ", c[i]);
  return 0;
}
