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 N = 1e5 + 5;
int ans[N];
struct point {
  int x, y;
  point(int x, int y) : x(x), y(y) {}
  point() : x(0), y(0) {}
  point operator+(point p) { return point(x + p.x, y + p.y); }
  point operator-(point p) { return point(x - p.x, y - p.y); }
  long long operator!() { return x * (long long)x + y * (long long)y; }
} v[N];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> v[i].x >> v[i].y;
  int lst = 0;
  point sum;
  for (int i = 1; i < n; i++) {
    int j = i;
    long long s1 = min(!(sum + v[j]), !(sum - v[j]));
    long long s2 = min(!(sum + v[lst]), !(sum - v[lst]));
    if (s1 > s2) swap(j, lst);
    s1 = !(sum + v[j]);
    s2 = !(sum - v[j]);
    if (s1 < s2)
      ans[j] = 1, sum = sum + v[j];
    else
      ans[j] = -1, sum = sum - v[j];
  }
  long long s1 = !(sum + v[lst]);
  long long s2 = !(sum - v[lst]);
  if (s1 < s2)
    ans[lst] = 1;
  else
    ans[lst] = -1;
  for (int i = 0; i < n; i++) cout << ans[i] << ' ';
  cout << endl;
}