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 int MAXN = 100005;
int n, top;
struct point {
  long long x, y;
  int id;
  friend point operator+(const point &a, const point &b) {
    return (point){a.x + b.x, a.y + b.y, 0};
  }
  friend point operator-(const point &a, const point &b) {
    return (point){a.x - b.x, a.y - b.y, 0};
  }
  inline long long len() { return x * x + y * y; }
};
int rev[MAXN * 2], fa[MAXN * 2];
stack<point> vec;
void init() {
  for (int i = top; i >= 1; i--) {
    if (fa[i]) rev[i] ^= rev[fa[i]];
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  top = n;
  for (int i = 1; i <= n; i++) {
    int x, y;
    cin >> x >> y;
    vec.push((point){x, y, i});
  }
  while (vec.size() >= 3) {
    point a[3];
    for (int i = 0; i < 3; i++) a[i] = vec.top(), vec.pop();
    int flag = 0;
    for (int i = 0; i < 3 && !flag; i++)
      for (int j = 0; j < i && !flag; j++) {
        if ((a[i] + a[j]).len() <= max(a[i].len(), a[j].len())) {
          vec.push(a[3 - i - j]);
          vec.push((point){a[i].x + a[j].x, a[i].y + a[j].y, ++top});
          fa[a[i].id] = fa[a[j].id] = top;
          flag = 1;
        } else if ((a[i] - a[j]).len() <= max(a[i].len(), a[j].len())) {
          vec.push(a[3 - i - j]);
          vec.push((point){a[i].x - a[j].x, a[i].y - a[j].y, ++top});
          rev[a[j].id] ^= 1;
          fa[a[i].id] = fa[a[j].id] = top;
          flag = 1;
        }
      }
    assert(flag == 1);
  }
  if (vec.size() == 2) {
    point a = vec.top();
    vec.pop();
    point b = vec.top();
    if ((a + b).len() >= (a - b).len()) rev[b.id] ^= 1;
  }
  init();
  point sum = (point){0, 0, 0};
  for (int i = 1; i <= n; i++) cout << (rev[i] ? -1 : 1) << " ";
  cout << endl;
  return 0;
}