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;
unsigned long long d_2 = 2250000000000;
struct V {
  long long x, y;
};
unsigned long long r(V v) { return v.x * v.x + v.y * v.y; }
int main() {
  int n;
  cin >> n;
  V v[n];
  for (int i = 0; i < n; ++i) {
    cin >> v[i].x >> v[i].y;
  }
  int c[n];
  V pos = {0, 0};
  unsigned long long p = 0;
  for (int i = 0; i < n; ++i) {
    unsigned long long d1 = r({pos.x + v[i].x, pos.y + v[i].y});
    unsigned long long d2 = r({pos.x - v[i].x, pos.y - v[i].y});
    if (d1 <= d2) {
      c[i] = 1;
      pos.x = pos.x + v[i].x;
      pos.y = pos.y + v[i].y;
    } else {
      c[i] = -1;
      pos.x = pos.x - v[i].x;
      pos.y = pos.y - v[i].y;
    }
    p = r(pos);
    int j = 0;
    while (p > d_2 && j < i) {
      unsigned long long dj =
          r({pos.x - 2 * c[j] * v[j].x, pos.y - 2 * c[j] * v[j].y});
      if (dj < p) {
        pos = {pos.x - 2 * c[j] * v[j].x, pos.y - 2 * c[j] * v[j].y};
        p = dj;
        c[j] *= -1;
      }
      ++j;
    }
  }
  for (int i = 0; i < n; ++i) cout << c[i] << " ";
  cout << endl;
}
