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 = 1000000007;
const int N = 100005;
const int M = 1000005;
int main() {
  int n;
  scanf("%d", &n);
  int x = 0, y = 0;
  vector<pair<int, int> > p;
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    scanf("%d", &y);
    p.push_back(make_pair(x, y));
  }
  x = y = 0;
  int ans[n];
  for (int i = n - 1; i >= 0; i--) {
    int x1 = p[i].first, y1 = p[i].second;
    long long xx = x + x1, yy = y + y1;
    long long xxx = x - x1, yyy = y - y1;
    long long d1 = xx * xx + yy * yy;
    long long d2 = xxx * xxx + yyy * yyy;
    if (d1 <= d2) {
      ans[i] = 1;
      x = xx;
      y = yy;
    } else {
      ans[i] = -1;
      x = xxx;
      y = yyy;
    }
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  return 0;
}