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 maxn = 100010;
long long X, Y;
int ans[maxn];
pair<pair<int, int>, int> s[maxn];
int main() {
  int N, x, y, i;
  scanf("%d", &N);
  for (i = 1; i <= N; i++) {
    scanf("%d%d", &x, &y);
    s[i] = make_pair(make_pair(x, y), i);
  }
  while (true) {
    random_shuffle(s + 1, s + N + 1);
    X = Y = 0;
    for (i = 1; i <= N; i++) {
      x = s[i].first.first;
      y = s[i].first.second;
      if ((long long)(X + x) * (X + x) + (Y + y) * (Y + y) <
          (long long)(X - x) * (X - x) + (Y - y) * (Y - y))
        X += x, Y += y, ans[s[i].second] = 1;
      else
        X -= x, Y -= y, ans[s[i].second] = -1;
    }
    if (X * X + Y * Y <= 2250000000000ll) {
      for (i = 1; i <= N; i++) printf("%d ", ans[i]);
      return 0;
    }
  }
  return 0;
}