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:200000')
using namespace std;
struct Point {
  long long x, y;
  Point() {
    x = 0;
    y = 0;
  };
  Point(long long _x, long long _y) {
    x = _x;
    y = _y;
  };
  void read() { scanf("%lld %lld", &x, &y); }
  Point operator+(Point o) const { return Point(x + o.x, y + o.y); }
  Point operator-(Point o) const { return Point(x - o.x, y - o.y); }
  long long norm() { return x * x + y * y; }
};
const int N = 100010;
const long long SAFE = 1500000LL * 1500000LL;
int n;
int s[N];
Point r[N];
int main() {
  srand(time(NULL));
  scanf("%d", &n);
  Point p;
  for (int i = 0; i < n; ++i) {
    Point q;
    q.read();
    r[i] = q;
    Point plus = p + q;
    Point minus = p - q;
    if (plus.norm() < minus.norm()) {
      p = plus;
      s[i] = 1;
    } else {
      p = minus;
      s[i] = -1;
    }
  }
  while (p.norm() > SAFE) {
    int idx = rand() % n;
    Point ax;
    if (s[idx] == 1) {
      ax = p - r[idx] - r[idx];
    } else {
      ax = p + r[idx] + r[idx];
    }
    if (ax.norm() < p.norm()) {
      s[idx] *= -1;
      p = ax;
    }
  }
  for (int i = 0; i < n; ++i) {
    printf("%d ", s[i]);
  }
  printf("\n");
  return 0;
}