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 long long maxn = 2.25e12;
struct data {
  int p, q, idx;
  data(int p, int q, int idx) : p(p), q(q), idx(idx) {}
  data() {}
  bool operator<(data d) const { return p < d.p; }
};
int ans[100010];
long long val(long long x, long long y) { return (x * x) + (y * y); }
int main(int argc, char const* argv[]) {
  srand(time(NULL));
  int n;
  scanf("%d", &n);
  vector<data> v;
  for (int i = 0; i < n; i++) {
    int p, q;
    scanf("%d %d", &p, &q);
    v.emplace_back(p, q, i);
  }
  sort(v.begin(), v.end());
  int cnt = 0;
  while (true) {
    long long x = 0;
    long long y = 0;
    for (auto i : v) {
      int p = i.p;
      int q = i.q;
      if (val(x + p, y + q) < val(x - p, y - q)) {
        x += p;
        y += q;
        ans[i.idx] = 1;
      } else {
        x -= p;
        y -= q;
        ans[i.idx] = -1;
      }
    }
    if (val(x, y) <= maxn) {
      for (int i = 0; i < n; i++) {
        printf("%d ", ans[i]);
      }
      printf("\n");
      exit(0);
    }
    random_shuffle(v.begin(), v.end());
    cnt++;
    assert(cnt <= 10);
  }
  return 0;
}