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 
#include <bits/stdc++.h>
using namespace std;
const long long int MAXL = 1.5e6, MAXL2 = MAXL * MAXL;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> x(n), y(n);
  for (int i = 0; i < (int)(n); i++) cin >> x[i] >> y[i];
  int iter = 100;
  while (iter--) {
    long long int xx = 0, yy = 0;
    vector<int> p(n);
    iota((p).begin(), (p).end(), 0);
    random_shuffle((p).begin(), (p).end());
    vector<int> ans(n);
    for (int i = 0; i < (int)(n); i++) {
      auto x1 = xx + x[p[i]], y1 = yy + y[p[i]];
      auto x2 = xx - x[p[i]], y2 = yy - y[p[i]];
      if (x1 * x1 + y1 * y1 < x2 * x2 + y2 * y2) {
        tie(xx, yy) = make_pair(x1, y1);
        ans[p[i]] = 1;
      } else {
        tie(xx, yy) = make_pair(x2, y2);
        ans[p[i]] = -1;
      }
    }
    if (xx * xx + yy * yy < MAXL2) {
      cerr << iter << ' ' << xx << ' ' << yy << ' '
           << double(xx * xx + yy * yy) / MAXL2 << endl;
      for (auto s : ans) cout << s << ' ';
      break;
    }
  }
  return 0;
}
