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;
long long int sol = 1500000;
long long int dist(long long int x, long long int y) { return x * x + y * y; }
bool comp(pair<pair<long long int, long long int>, long long int>& a,
          pair<pair<long long int, long long int>, long long int>& b) {
  return dist(a.first.first, a.first.second) >
         dist(b.first.first, b.first.second);
}
int main() {
  int n;
  cin >> n;
  vector<pair<pair<long long int, long long int>, long long int> > V(n);
  for (long long int i = 0; i < n; ++i) {
    long long int x;
    long long int y;
    cin >> x >> y;
    V[i] = make_pair(make_pair(x, y), i);
  }
  sort(V.begin(), V.end(), comp);
  bool end = false;
  vector<long long int> M(n);
  vector<long long int> MM(n);
  long long int x0 = 0;
  long long int y0 = 0;
  for (long long int i = 0; i < n; ++i) {
    long long int x = V[i].first.first;
    long long int y = V[i].first.second;
    long long int s1 = dist(x + x0, y + y0);
    long long int s2 = dist(x0 - x, y0 - y);
    if (s1 <= s2) {
      x0 = x + x0;
      y0 = y + y0;
      M[V[i].second] = 1;
      MM[i] = V[i].second;
    } else {
      x0 = x0 - x;
      y0 = y0 - y;
      M[V[i].second] = 0;
      MM[i] = V[i].second;
    }
  }
  int c = 0;
  int d = 0;
  while (not end) {
    if (x0 * x0 + y0 * y0 <= sol * sol) end = true;
    if (not end) {
      M[V[d].second] = 1 - M[V[d].second];
      if (M[V[d].second] == 0) {
        x0 -= 2 * V[d].first.first;
        y0 -= 2 * V[d].first.second;
      } else {
        x0 += 2 * V[d].first.first;
        y0 += 2 * V[d].first.second;
      }
      if (d == 0) {
        ++c;
        d = c;
      } else
        --d;
    }
  }
  for (int i = 0; i < n; ++i) {
    if (i) cout << " ";
    if (M[i])
      cout << "1";
    else
      cout << "-1";
  }
  cout << endl;
}