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;
long long maxi = LLONG_MIN;
long long mini = LLONG_MAX;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
long long t, i, j, k, n, temp, flag, m;
struct pt {
  long double x, y;
  long long pos;
};
int main() {
  cin >> n;
  pt a[n];
  vector<long long> ans(n);
  for (i = 0; i < n; i++) {
    cin >> a[i].x >> a[i].y;
    a[i].pos = i;
  }
  long double rpd = (1.5 * 1000000) * (1.5 * 1000000);
  while (true) {
    long double px = 0, py = 0, pd;
    for (i = 0; i < n; i++) {
      long double px1 = px + a[i].x;
      long double py1 = py + a[i].y;
      long double px2 = px - a[i].x;
      long double py2 = py - a[i].y;
      long double pd1 = (px1 * px1) + (py1 * py1);
      long double pd2 = (px2 * px2) + (py2 * py2);
      if (pd1 <= pd2) {
        px = px1;
        py = py1;
        pd = pd1;
        ans[a[i].pos] = 1;
      } else {
        px = px2;
        py = py2;
        pd = pd2;
        ans[a[i].pos] = -1;
      }
    }
    if (pd <= rpd) break;
    random_shuffle(a, a + n);
  }
  for (i = 0; i < n; i++) {
    cout << ans[i] << ' ';
  }
  return 0;
}
