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 N = 1e5 + 10;
int n, dx[N], dy[N], answer[N];
long long dist(int x, int y) { return (long long)x * x + (long long)y * y; }
bool better(pair<int, int> o, int x, int y) {
  return dist(o.first + x, o.second + y) <= dist(o.first - x, o.second - y);
}
int main() {
  scanf("%d", &n);
  pair<int, int> vec(0, 0);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &dx[i], &dy[i]);
    if (better(vec, dx[i], dy[i])) {
      vec.first += dx[i];
      vec.second += dy[i];
      answer[i] = 1;
    } else {
      vec.first -= dx[i];
      vec.second -= dy[i];
      answer[i] = -1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    vec.first -= answer[i] * dx[i];
    vec.second -= answer[i] * dy[i];
    if (better(vec, -answer[i] * dx[i], -answer[i] * dy[i])) {
      vec.first -= answer[i] * dx[i];
      vec.second -= answer[i] * dy[i];
      answer[i] = -answer[i];
    } else {
      vec.first += answer[i] * dx[i];
      vec.second += answer[i] * dy[i];
    }
  }
  for (int i = 1; i <= n; ++i) {
    printf("%d%c", answer[i], " \n"[i == n]);
  }
  return 0;
}