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;
long long lim = 1500000LL * 1500000LL;
int main() {
  long long n;
  scanf("%lld", &n);
  pair<long long, pair<long long, long long> > arr[n];
  for (long long i = 0; i < n; i++) {
    scanf("%lld%lld", &(arr[i].first), &(arr[i].second.first));
    arr[i].second.second = i;
  }
  int res[n];
  while (1) {
    long long c1 = 0, c2 = 0;
    random_shuffle(arr, arr + n);
    for (long long i = 0; i < n; i++) {
      if ((abs(c1 - arr[i].first) + abs(c2 - arr[i].second.first)) <
          (abs(c1 + arr[i].first) + abs(c2 + arr[i].second.first))) {
        res[arr[i].second.second] = -1;
        c1 -= arr[i].first;
        c2 -= arr[i].second.first;
      } else {
        res[arr[i].second.second] = 1;
        c1 += arr[i].first;
        c2 += arr[i].second.first;
      }
    }
    if ((c1 * c1 + c2 * c2) <= lim) break;
  }
  for (long long i = 0; i < n; i++) {
    printf("%d  ", res[i]);
  }
  return 0;
}