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 + 5;
const int R = 1500000;
const double M = 1.0 * R * R;
int n, x[N], y[N], v[N];
double func(long long x, long long y) { return 1.0 * x * x + 1.0 * y * y; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> x[i] >> y[i];
  long long first = 0ll, second = 0ll;
  for (int i = 1; i <= n; i++) {
    double r1 = func(first + x[i], second + y[i]);
    double r2 = func(first - x[i], second - y[i]);
    if (r1 < r2)
      v[i] = 1;
    else
      v[i] = -1;
    first += x[i] * v[i];
    second += y[i] * v[i];
  }
  while (func(first, second) > M) {
    for (int i = 1; i <= n; i++) {
      double r1 = func(first, second);
      double r2 = func(first - 2 * v[i] * x[i], second - 2 * v[i] * y[i]);
      if (r2 < r1) {
        first -= 2 * v[i] * x[i];
        second -= 2 * v[i] * y[i];
        v[i] = -v[i];
      }
    }
  }
  for (int i = 1; i <= n; i++) cout << v[i] << " ";
  printf("\n");
  return 0;
}