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;
int n, x, y;
vector<pair<pair<int, int>, int> > v;
int cev[100005];
double euc(pair<int, int> p) {
  return sqrt(1ll * p.first * p.first + 1ll * p.second * p.second);
}
void solve() {
  vector<int> ans;
  pair<int, int> po = {0, 0};
  for (pair<pair<int, int>, int> vv : v) {
    pair<int, int> p1 = {po.first + vv.first.first,
                         po.second + vv.first.second};
    pair<int, int> p2 = {po.first - vv.first.first,
                         po.second - vv.first.second};
    if (euc(p1) < euc(p2)) {
      ans.push_back(1);
      po = p1;
    } else {
      ans.push_back(-1);
      po = p2;
    }
  }
  if (euc(po) > 1500000) return;
  for (int i = 0; i < (int)ans.size(); i++) cev[v[i].second] = ans[i];
  for (int i = 1; i <= n; i++) printf("%d ", cev[i]);
  exit(0);
}
int main() {
  srand(time(NULL));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &x, &y);
    v.push_back({{x, y}, i});
  }
  while (1) {
    solve();
    random_shuffle(v.begin(), v.end());
  }
}