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;
const int N = 112345;
int n;
pair<long long, long long> v[N];
int ans[N];
int main(void) {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second;
  }
  long long cx = 0, cy = 0;
  for (int i = 0; i < n; i++) {
    long long x = cx + v[i].first;
    long long y = cy + v[i].second;
    long long a1 = x * x + y * y;
    x = cx - v[i].first;
    y = cy - v[i].second;
    long long a2 = x * x + y * y;
    if (a1 < a2) {
      cx += v[i].first;
      cy += v[i].second;
      ans[i] = 1;
    } else {
      cx -= v[i].first;
      cy -= v[i].second;
      ans[i] = -1;
    }
  }
  if (cx * cx + cy * cy > 2250000000000) {
    for (int i = 0; i < n; i++) {
      long long nx = cx, ny = cy;
      if (ans[i] == 1) {
        nx -= 2 * v[i].first;
        ny -= 2 * v[i].second;
      } else {
        nx += 2 * v[i].first;
        ny += 2 * v[i].second;
      }
      if (nx * nx + ny * ny <= 2250000000000) {
        ans[i] *= -1;
        break;
      }
      if (i == n - 1) assert(false);
    }
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
