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;
bool cmp(pair<complex<double>, int> p1, pair<complex<double>, int> p2) {
  if (abs(p1.first) > abs(p2.first)) return true;
  return false;
}
bool cmp2(pair<complex<double>, int> p1, pair<complex<double>, int> p2) {
  if (abs(real(p1.first)) > abs(real(p2.first))) return true;
  return false;
}
bool cmp3(pair<complex<double>, int> p1, pair<complex<double>, int> p2) {
  if (abs(imag(p1.first)) > abs(imag(p2.first))) return true;
  return false;
}
int main() {
  int i, n;
  long long x, y;
  vector<pair<complex<double>, int> > points;
  vector<long long> c;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x >> y;
    points.push_back(make_pair(complex<double>(x, y), i));
    c.push_back(0);
  }
  sort(points.begin(), points.end(), cmp);
  c[points[0].second] = 1;
  complex<double> sum = complex<double>(0, 0);
  sum += points[0].first;
  for (i = 1; i < n; i++) {
    complex<double> sum1 = sum + points[i].first;
    complex<double> sum2 = sum - points[i].first;
    if (abs(sum1) > abs(sum2)) {
      c[points[i].second] = -1;
      sum = sum2;
    } else {
      c[points[i].second] = 1;
      sum = sum1;
    }
  }
  if (abs(sum) > 1500000) {
    sort(points.begin(), points.end(), cmp2);
    c[points[0].second] = 1;
    sum = complex<double>(0, 0);
    sum += points[0].first;
    for (i = 1; i < n; i++) {
      complex<double> sum1 = sum + points[i].first;
      complex<double> sum2 = sum - points[i].first;
      if (abs(sum1) > abs(sum2)) {
        c[points[i].second] = -1;
        sum = sum2;
      } else {
        c[points[i].second] = 1;
        sum = sum1;
      }
    }
    if (abs(sum) > 1500000) {
      sort(points.begin(), points.end(), cmp3);
      c[points[0].second] = 1;
      sum = complex<double>(0, 0);
      sum += points[0].first;
      for (i = 1; i < n; i++) {
        complex<double> sum1 = sum + points[i].first;
        complex<double> sum2 = sum - points[i].first;
        if (abs(sum1) > abs(sum2)) {
          c[points[i].second] = -1;
          sum = sum2;
        } else {
          c[points[i].second] = 1;
          sum = sum1;
        }
      }
      for (i = 0; i < n; i++) {
        cout << c[i] << " ";
      }
    } else {
      for (i = 0; i < n; i++) {
        cout << c[i] << " ";
      }
    }
  } else {
    for (i = 0; i < n; i++) {
      cout << c[i] << " ";
    }
  }
}