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;
int main() {
  ios::sync_with_stdio(false);
  int n, a, b;
  long long x, y, X[3], Y[3], g;
  cin >> n;
  int P[n];
  vector<int> V[3];
  if (n == 1) {
    cout << 1;
    return 0;
  }
  for (int i = 0; i < 2; i++) {
    cin >> x >> y;
    V[i].push_back(i);
    X[i] = x;
    Y[i] = y;
    P[i] = 1;
  }
  for (int i = 2; i < n; i++) {
    cin >> x >> y;
    g = 1;
    for (int u = 0; u < 2; u++) {
      if (sqrt((x + X[u]) * (x + X[u]) + (y + Y[u]) * (y + Y[u])) <= 1000000 &&
          g == 1) {
        X[u] += x;
        Y[u] += y;
        V[u].push_back(i);
        P[i] = 1;
        g = 0;
      }
      if (sqrt((x - X[u]) * (x - X[u]) + (y - Y[u]) * (y - Y[u])) <= 1000000 &&
          g == 1) {
        X[u] -= x;
        Y[u] -= y;
        V[u].push_back(i);
        P[i] = -1;
        g = 0;
      }
    }
    if (g == 1) {
      P[i] = 1;
      V[2].clear();
      V[2].push_back(i);
      a = 0;
      b = 1;
      if (V[1].size() > V[0].size()) {
        swap(a, b);
      }
      if (sqrt((X[0] - X[1]) * (X[0] - X[1]) + (Y[0] - Y[1]) * (Y[0] - Y[1])) <=
          1000000) {
        for (int u = 0; u < V[b].size(); u++) {
          P[V[b][u]] *= -1;
        }
        X[a] -= X[b];
        Y[a] -= Y[b];
      } else {
        X[a] += X[b];
        Y[a] += Y[b];
      }
      for (int u = 0; u < V[b].size(); u++) {
        V[a].push_back(V[b][u]);
      }
      X[b] = x;
      Y[b] = y;
      V[b].clear();
      V[b] = V[2];
    }
  }
  if (sqrt((X[0] - X[1]) * (X[0] - X[1]) + (Y[0] - Y[1]) * (Y[0] - Y[1])) <=
      1500000) {
    for (int i = 0; i < V[1].size(); i++) {
      P[V[1][i]] *= -1;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << P[i] << " ";
  }
  return 0;
}