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>
#pragma GCC optimize("O3")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long mod = 1000000007;
long long r(pair<int, int> a) {
  long long x = a.first;
  long long y = a.second;
  return x * x + y * y;
}
pair<int, int> operator+(pair<int, int> a, pair<int, int> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
pair<int, int> operator-(pair<int, int> a, pair<int, int> b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<pair<pair<int, int>, vector<int>>> a;
  for (int i = 1; i <= N; i++) {
    int x, y;
    cin >> x >> y;
    a.push_back({{x, y}, {i}});
  }
  shuffle(a.begin(), a.end(), rng);
  while (a.size() >= 3) {
    int i = a.size() - 1;
    int j = a.size() - 2;
    int wj = 1;
    pair<int, int> mnn = {1e8, 1e8};
    for (int x = a.size() - 3; x < (int)a.size(); x++) {
      for (int y = x + 1; y < (int)a.size(); y++) {
        if (x == y) continue;
        pair<int, int> v1 = a[x].first + a[y].first;
        pair<int, int> v2 = a[x].first - a[y].first;
        if (r(v1) < r(mnn)) {
          mnn = v1;
          i = x;
          j = y;
          wj = 1;
        }
        if (r(v2) < r(mnn)) {
          mnn = v2;
          i = x;
          j = y;
          wj = -1;
        }
      }
    }
    a[i].first = a[j].first = mnn;
    if (a[j].second.size() < a[i].second.size()) {
      for (int c : a[j].second) {
        a[i].second.push_back(c * wj);
      }
      a.erase(a.begin() + j);
    } else {
      if (wj == 1) {
        for (int c : a[i].second) {
          a[j].second.push_back(c * wj);
        }
        a.erase(a.begin() + i);
      } else {
        a[j].first = make_pair(0, 0) - a[j].first;
        for (int c : a[i].second) {
          a[j].second.push_back(-c);
        }
        a.erase(a.begin() + i);
      }
    }
  }
  if (a.size() == 2) {
    if (r(a[0].first - a[1].first) < r(a[0].first + a[1].first)) {
      for (int &i : a[1].second) i *= -1;
    }
  }
  int w[N + 1];
  for (auto ii : a) {
    for (int i : ii.second) {
      if (i > 0)
        w[i] = 1;
      else
        w[-i] = -1;
    }
  }
  for (int i = 1; i <= N; i++) cout << w[i] << " ";
}