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;
const double EPS = 1e-2;
const double PI = 3.14159265;
struct pt {
  long long x, y;
  int id;
  double ang;
  pt(int _x = 0, int _y = 0, int _id = -1) : x(_x), y(_y), id(_id) {
    ang = atan2(y, x);
  }
};
long long score(long long a, long long b) { return a * a + b * b; }
bool cmp(const pt& a, const pt& b) {
  if (a.ang == b.ang) return abs(a.x) + abs(a.y) < abs(b.x) + abs(b.y);
  return a.ang < b.ang;
}
int main() {
  srand(15061994);
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<pt> X;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    X.push_back(pt(x, y, i));
  }
  while (true) {
    random_shuffle(X.begin(), X.end());
    vector<int> ret(n);
    pt d;
    for (int i = 0; i < n; i++) {
      if (score(d.x + X[i].x, d.y + X[i].y) <=
          score(d.x - X[i].x, d.y - X[i].y)) {
        ret[X[i].id] = 1;
        d.x += X[i].x;
        d.y += X[i].y;
      } else {
        ret[X[i].id] = -1;
        d.x -= X[i].x;
        d.y -= X[i].y;
      }
    }
    if (score(d.x, d.y) <= 1.5e6 * 1.5e6) {
      for (int i = 0; i < n; i++) cout << ret[i] << " ";
      cout << "\n";
      break;
    }
  }
  return 0;
}