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 M = 1e5 + 10;
int n;
long long Sqr(long long a) { return a * a; }
struct Vector {
  long long x, y, id;
  void in(int i) {
    scanf("%I64d%I64d", &x, &y);
    id = i;
  }
  Vector() {}
  Vector(long long a, long long b) : x(a), y(b) {}
  Vector operator+(const Vector &a) const { return Vector(x + a.x, y + a.y); }
  long long operator!() { return Sqr(x) + Sqr(y); }
  Vector operator~() { return Vector(-x, -y); }
  bool operator<(const Vector &a) const {
    return x < a.x || (x == a.x && y < a.y);
  }
} V[M], now, t1, t2, t3, t4, tt1, tt2;
long long moc(Vector a) { return sqrt(!a); }
int ans[M], v1, v2;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) V[i].in(i);
  sort(V + 1, V + n + 1);
  for (int i = 1, j = n; i <= j;) {
    t1 = now + V[i];
    t2 = now + (~V[i]);
    t3 = now + V[j];
    t4 = now + (~V[j]);
    if ((!t1) < (!t2))
      tt1 = t1, v1 = 1;
    else
      tt1 = t2, v1 = -1;
    if ((!t3) < (!t4))
      tt2 = t3, v2 = 1;
    else
      tt2 = t4, v2 = -1;
    if ((!tt1) < (!tt2))
      now = tt1, ans[V[i].id] = v1, ++i;
    else
      now = tt2, ans[V[j].id] = v2, --j;
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  puts("");
  return 0;
}