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 nmax = 1e5;
short int semn[4 * nmax + 1], rasp[nmax + 1];
pair<pair<int, int>, int> v[nmax + 1];
long long lg(pair<int, int> a) {
  return 1LL * a.first * a.first + 1LL * a.second * a.second;
}
pair<int, int> operator+(pair<int, int> a, pair<int, int> b) {
  return {a.first + b.first, a.second + b.second};
}
pair<int, int> operator-(pair<int, int> a, pair<int, int> b) {
  return {a.first - b.first, a.second - b.second};
}
pair<int, int> solve(int nod, int x, int y) {
  semn[nod] = 1;
  if (x == y) {
    return v[x].first;
  }
  int mij = (x + y) / 2;
  pair<int, int> st, dr;
  st = solve(2 * nod, x, mij);
  dr = solve(2 * nod + 1, mij + 1, y);
  if (lg(st + dr) < lg(st - dr)) {
    return st + dr;
  } else {
    semn[2 * nod + 1] *= -1;
    return st - dr;
  }
  return {-1, -1};
}
void dfs(int nod, int x, int y) {
  if (x == y) {
    rasp[v[x].second] = semn[nod];
    return;
  }
  int mij = (x + y) / 2;
  semn[2 * nod] *= semn[nod];
  semn[2 * nod + 1] *= semn[nod];
  dfs(2 * nod, x, mij);
  dfs(2 * nod + 1, mij + 1, y);
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> v[i].first.first >> v[i].first.second;
    v[i].second = i;
  }
  srand(time(0));
  while (1) {
    random_shuffle(v + 1, v + n + 1);
    pair<int, int> vct = solve(1, 1, n);
    if (sqrt(lg(vct)) <= 1e6 * 1.5) {
      dfs(1, 1, n);
      for (int i = 1; i <= n; ++i) {
        cout << rasp[i] << " ";
      }
      cout << "\n";
      break;
    }
  }
  return 0;
}