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:102400000,102400000")
using namespace std;
const int N = 3e5 + 10;
const long long pp = 1500000;
long long gcd(long long p, long long q) { return q == 0 ? p : gcd(q, p % q); }
int to[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
int n;
struct nd {
  long long x, y;
  int id;
} p[N];
int a[N];
long long d(long long a, long long b) { return a * a + b * b; }
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].x >> p[i].y;
    p[i].id = i;
  }
  while (1) {
    random_shuffle(p + 1, p + n + 1);
    long long x = 0, y = 0;
    for (int i = 1; i <= n; i++) {
      if (d(x - p[i].x, y - p[i].y) < d(x + p[i].x, y + p[i].y)) {
        x -= p[i].x, y -= p[i].y;
        a[p[i].id] = -1;
      } else {
        x += p[i].x, y += p[i].y;
        a[p[i].id] = 1;
      }
    }
    if (x * x + y * y <= pp * pp) {
      for (int i = 1; i <= n; i++) {
        cout << a[i];
        if (i != n) cout << " ";
      }
      break;
    }
  }
  return 0;
}