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;
vector<int> adj[100001];
bool visit[100001];
const int mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, i;
  cin >> n;
  vector<int> v;
  long long x = 0, y = 0;
  long long p[100001], q[100001];
  long long a, b, c, d;
  for (long long i = 0; i < n; i += 1) cin >> p[i] >> q[i];
  for (long long j = n - 1; j >= 0; j -= 1) {
    a = p[j], b = q[j];
    if (x >= 0 && y >= 0) {
      if (a >= 0 && b >= 0)
        x -= a, y -= b, v.push_back(-1);
      else if (a <= 0 && b <= 0)
        x += a, y += b, v.push_back(1);
      else {
        c = x + a, d = y + b;
        double a1 = sqrt(c * c + d * d);
        c = x - a, d = y - b;
        double a2 = sqrt(c * c + d * d);
        if (a1 < a2)
          x += a, y += b, v.push_back(1);
        else
          x -= a, y -= b, v.push_back(-1);
      }
    } else if (x <= 0 && y <= 0) {
      if (a >= 0 && b >= 0)
        x += a, y += b, v.push_back(1);
      else if (a <= 0 && b <= 0)
        x -= a, y -= b, v.push_back(-1);
      else {
        c = x + a, d = y + b;
        double a1 = sqrt(c * c + d * d);
        c = x - a, d = y - b;
        double a2 = sqrt(c * c + d * d);
        if (a1 < a2)
          x += a, y += b, v.push_back(1);
        else
          x -= a, y -= b, v.push_back(-1);
      }
    } else {
      c = x + a, d = y + b;
      double a1 = sqrt(c * c + d * d);
      c = x - a, d = y - b;
      double a2 = sqrt(c * c + d * d);
      if (a1 < a2)
        x += a, y += b, v.push_back(1);
      else
        x -= a, y -= b, v.push_back(-1);
    }
  }
  reverse(v.begin(), v.end());
  for (long long i = 0; i < v.size(); i += 1) cout << v[i] << " ";
  return 0;
}