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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const long long mod = 1e9 + 7;
const int maxn = 1e6 + 5;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct coord {
  long long i, x, y;
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<coord> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].x >> v[i].y;
    v[i].i = i;
  }
  while (true) {
    shuffle(v.begin(), v.end(), rng);
    long long x = 0, y = 0;
    vector<int> res(n);
    for (auto co : v) {
      long long x1 = x + co.x;
      long long y1 = y + co.y;
      long long x2 = x - co.x;
      long long y2 = y - co.y;
      if (x1 * x1 + y1 * y1 < x2 * x2 + y2 * y2) {
        res[co.i] = +1;
        x = x1;
        y = y1;
      } else {
        res[co.i] = -1;
        x = x2;
        y = y2;
      }
    }
    const long long dist = 1.5e6;
    if (x * x + y * y <= dist * dist) {
      for (int i : res) cout << i << " ";
      cout << endl;
      return 0;
    }
  }
  assert(0);
  return 0;
}