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 GCC optimize("Ofast")
using namespace std;
const long long N = 1e5 + 1, MAX = 1.5 * 1e6;
mt19937 rng(chrono::high_resolution_clock().now().time_since_epoch().count());
pair<long long, long long> add(const pair<long long, long long> &a,
                               const pair<long long, long long> &b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
bool ok(const pair<long long, long long> &a) {
  if (a.first > MAX || a.second > MAX)
    return false;
  else
    return a.first * a.first + a.second * a.second <= MAX * MAX;
}
long long n, c[N];
pair<long long, long long> v[N];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; ++i) cin >> v[i].first >> v[i].second;
  for (long long t = 1; t <= 2e8 / n; ++t) {
    pair<long long, long long> now = {0, 0};
    for (long long i = 1; i <= n; ++i) {
      c[i] = (rng() % 2 == 0 ? -1 : 1);
      if (!ok(add(now, make_pair(c[i] * v[i].first, c[i] * v[i].second))))
        c[i] = -c[i];
      now = add(now, make_pair(c[i] * v[i].first, c[i] * v[i].second));
    }
    if (ok(now)) break;
  }
  for (long long i = 1; i <= n; ++i) cout << c[i] << ' ';
}