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;
const int INF = 1000000000;
const long long LINF = 1LL * INF * INF;
const int MOD = INF + 7;
vector<pair<int, pair<long long, long long> > > s[2];
int resp[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  std::srand(unsigned(std::time(0)));
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    if (true) {
      s[0].push_back(make_pair(i, make_pair(x, y)));
    } else {
      s[1].push_back(make_pair(i, make_pair(x, y)));
    }
  }
  random_shuffle(s[0].begin(), s[0].end());
  random_shuffle(s[1].begin(), s[1].end());
  int a = 0, b = 0;
  for (int k = 0; k < 2; k++) {
    for (int i = 0; i < (int)s[k].size(); i++) {
      long long na1 = a + s[k][i].second.first, nb1 = b + s[k][i].second.second;
      long long na2 = a - s[k][i].second.first, nb2 = b - s[k][i].second.second;
      long long v1 = na1 * na1 + nb1 * nb1, v2 = na2 * na2 + nb2 * nb2;
      if (v1 > v2) {
        resp[s[k][i].first] = -1;
        a -= s[k][i].second.first;
        b -= s[k][i].second.second;
      } else {
        resp[s[k][i].first] = 1;
        a += s[k][i].second.first;
        b += s[k][i].second.second;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    cout << resp[i] << " ";
  }
  cout << endl;
  cerr << sqrt(1. * a * a + 1. * b * b) << endl;
}