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;
struct pt {
  int x, y;
  pt(int x, int y) : x(x), y(y) {}
  pt() : x(0), y(0) {}
  pt operator=(pt p) {
    x = p.x, y = p.y;
    return pt(x, y);
  }
  pt operator+=(pt p) {
    x += p.x, y += p.y;
    return pt(x, y);
  }
  pt operator-=(pt p) {
    x -= p.x, y -= p.y;
    return pt(x, y);
  }
};
double dot(pt a, pt b) { return a.x * (double)b.x + a.y * (double)b.y; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  int n = 0, a, b, ind[100005], u, rev[100005];
  pt p[100005];
  double v[100005];
  set<pair<int, int> > ans;
  cin >> u;
  for (int i = 0; i < u; i++) {
    cin >> a >> b;
    if (a == 0 && b == 0)
      ans.insert(pair<int, int>(i, 1));
    else
      p[n] = pt(a, b), ind[n] = n, rev[n] = i, n++, v[i] = atan2(b, a);
  }
  sort(ind, ind + n, [&](int i, int j) { return v[i] < v[j]; });
  pt res = pt(0, 0);
  for (int i = 0; i < n; i++) {
    double l = dot(res, p[ind[i]]);
    if (l >= 0)
      res -= p[ind[i]], ans.insert(pair<int, int>(rev[ind[i]], -1));
    else
      res += p[ind[i]], ans.insert(pair<int, int>(rev[ind[i]], 1));
  }
  for (auto x : ans) cout << x.second << " ";
  return 0;
}
