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 int mod = 1e9 + 7;
const int N = 1e6 + 10;
const long long INF = 1e18;
const long double EPS = 1e-12;
vector<pair<pair<int, int>, int> > vec;
int c[100005];
int main() {
  srand(time(NULL));
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (int)n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    vec.push_back(make_pair(make_pair(x, y), i));
  }
  long long tmp = 1.5 * (long long)(1e6);
  tmp *= tmp;
  for (int iii = 0; iii < (int)100; iii++) {
    random_shuffle(vec.begin(), vec.end());
    long long x = 0;
    long long y = 0;
    for (int i = 0; i < (int)n; i++) {
      int id = vec[i].second;
      long long x_ = x + vec[i].first.first;
      long long y_ = y + vec[i].first.second;
      long long _x = x - vec[i].first.first;
      long long _y = y - vec[i].first.second;
      if (x_ * x_ + y_ * y_ >= _x * _x + _y * _y) {
        c[id] = -1;
        x = _x;
        y = _y;
      } else {
        c[id] = 1;
        x = x_;
        y = y_;
      }
    }
    if (x * x + y * y <= tmp) {
      break;
    }
  }
  for (int i = 0; i < (int)n; i++) {
    printf("%d ", c[i]);
  }
  printf("\n");
  return 0;
}