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;
int N;
pair<int, int> V[100005];
int Res[100005];
pair<int, int> P[100005];
void Read() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d%d", &V[i].first, &V[i].second);
  }
}
long long dist(pair<int, int> v) {
  return 1LL * v.first * v.first + 1LL * v.second * v.second;
}
void Solve() {
  P[0] = make_pair(0, 0);
  for (int i = 1; i <= N; i++) {
    int k = i;
    for (k = i - 1; k >= 0; k--)
      if (dist(P[k]) <= 1000000000000) break;
    int nb = i - k, PMin = 0;
    long long Min = 10000000000000000;
    for (int conf = 0; conf < (1 << nb); conf++) {
      pair<int, int> p = P[k];
      for (int j = 0; j < nb; j++) {
        if ((conf & (1 << j)) != 0)
          p.first -= V[k + 1 + j].first, p.second -= V[k + 1 + j].second;
        else
          p.first += V[k + 1 + j].first, p.second += V[k + 1 + j].second;
      }
      if (Min > dist(p)) {
        Min = dist(p);
        PMin = conf;
      }
    }
    pair<int, int> p = P[k];
    for (int j = 0; j < nb; j++) {
      if ((PMin & (1 << j)) != 0)
        p.first -= V[k + 1 + j].first, p.second -= V[k + 1 + j].second,
            P[k + 1 + j] = p, Res[k + 1 + j] = -1;
      else
        p.first += V[k + 1 + j].first, p.second += V[k + 1 + j].second,
            P[k + 1 + j] = p, Res[k + 1 + j] = 1;
    }
  }
  for (int i = 1; i <= N; i++) printf("%d ", Res[i]);
  printf("\n");
}
int main() {
  Read();
  Solve();
  return 0;
}
