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>
#pragma comment(linker, "/STACK:64000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
const int INF = 2e9 + 179, MOD = 1e9 + 7;
const long long int BIGINF = 2e18 + 179;
const long double EPS = 0.000001;
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(20);
  srand(179);
  int n;
  cin >> n;
  vector<pair<long long int, long long int>> a(n);
  long long int px = 0, py = 0;
  vector<int> ans(n, 1);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first >> a[i].second;
    if (rand() % 2) {
      ans[i] = -1;
      px -= a[i].first;
      py -= a[i].second;
    } else {
      px += a[i].first;
      py += a[i].second;
    }
  }
  long long int urmom = 2250000000000;
  long long int urass = 1500000;
  int RenoInDeck = 1e8;
  while (RenoInDeck) {
    if (abs(px) < urass && abs(py) < urass) {
      if (px * px + py * py < urmom) {
        for (int i = 0; i < n; ++i) {
          cout << ans[i] << ' ';
        }
        cout << endl;
        return 0;
      }
    }
    int ind = rand() % n;
    if (ans[ind] == 1) {
      px -= 2 * a[ind].first;
      py -= 2 * a[ind].second;
      ans[ind] = -1;
    } else {
      px += 2 * a[ind].first;
      py += 2 * a[ind].second;
      ans[ind] = 1;
    }
    RenoInDeck--;
  }
  cout << "Skritiy pool" << endl;
  return 0;
}