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;
#pragma warning(disable : 4996)
int store[100001][3], ans[100001];
double dis(int x, int y) {
  long long temp = (long long)x * x + (long long)y * y;
  return sqrt(temp);
}
int fing(int x, int y, int &xi, int &yi) {
  long long a = (long long)(xi + x) * (xi + x) + (long long)(yi + y) * (yi + y);
  long long b = (long long)(xi - x) * (xi - x) + (long long)(yi - y) * (yi - y);
  if (a <= b) {
    xi += x, yi += y;
    return 1;
  }
  xi -= x;
  yi -= y;
  return -1;
}
int main() {
  int m, x, y, temx, temy;
  x = y = 0;
  scanf("%d", &m);
  for (int i = 0; i < m; i++)
    scanf("%d %d", &store[i][0], &store[i][1]), store[i][2] = i;
  while (1) {
    for (int i = 0; i < m; i++) {
      ans[store[i][2]] = fing(store[i][0], store[i][1], x, y);
    }
    if (dis(x, y) > 1500000)
      random_shuffle(store, store + m), x = 0, y = 0;
    else
      break;
  }
  for (int i = 0; i < m; i++) cout << ans[i] << " ";
  return 0;
}