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 maxn = 100005;
int n;
pair<int, int> a[maxn];
long long Max = 2250000000000;
vector<int> v;
void print() {
  for (auto i : v) cout << i << " ";
  exit(0);
}
map<pair<int, pair<int, int> >, bool> vis;
void first(int i, long long x, long long y) {
  if (i == n) {
    if (x * x + y * y <= Max) print();
    return;
  }
  if (vis[{i, {x, y}}]) return;
  vis[{i, {x, y}}] = true;
  if ((x - a[i].first) * (x - a[i].first) +
          (y - a[i].second) * (y - a[i].second) <=
      Max)
    v[i] = -1, first(i + 1, x - a[i].first, y - a[i].second);
  v[i] = 1, first(i + 1, x + a[i].first, y + a[i].second);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  cin >> n;
  v.resize(n);
  for (int i = 0; i < n; ++i) cin >> a[i].first >> a[i].second;
  first(0, 0, 0);
  return 0;
}