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 
#include <bits/stdc++.h>
const int N = (int)1e6 + 1;
const int N_ = (int)5e3 + 1;
const int LN = 20;
const long double PI = 3.1415926535897932384626433832795;
const long double EPS = 1e-8;
const int MOD = (int)1e9 + 7;
const long long LINF = (int64_t)1e15;
const int INF = (int)1e9;
const int DEL = (int)1e6 + 1;
const int PR = 115249;
const int CNT = 15626;
using namespace std;
template <typename T>
T sqr(T a) {
  return a * a;
}
int n;
pair<int, int> v[N];
void read() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> v[i].first >> v[i].second;
}
void solve() {
  long long x = 0, y = 0;
  int ans[N] = {};
  for (int i = n - 1; i >= 0; i--) {
    long long nx1 = x + v[i].first, ny1 = y + v[i].second;
    long long nx2 = x - v[i].first, ny2 = y - v[i].second;
    long long d1 = sqr(nx1) + sqr(ny1), d2 = sqr(nx2) + sqr(ny2);
    if (sqrt(d1 + .0) <= sqrt(d2 + .0)) {
      ans[i] = 1;
      x = nx1, y = ny1;
    } else {
      ans[i] = -1;
      x = nx2, y = ny2;
    }
  }
  for (int i = 0; i < n; i++) cout << ans[i] << ' ';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(20);
  read();
  solve();
}
