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 maxn = 100005;
struct Point {
  int id;
  long long x, y, d;
  Point(long long x, long long y) {
    this->x = x;
    this->y = y;
    this->d = x * x + y * y;
  }
  Point() {}
  inline bool operator<(const Point& other) const { return this->d < other.d; }
};
Point p[maxn];
inline Point operator+(Point a, Point b) { return Point(a.x + b.x, a.y + b.y); }
inline Point operator-(Point a, Point b) { return Point(a.x - b.x, a.y - b.y); }
inline bool cmp(Point x, Point y) { return x.d > y.d; }
int ans[maxn];
int anss[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    p[i] = Point(x, y);
    p[i].id = i;
  }
  Point lst = Point(0x3f3f3f3fLL, 0x3f3f3f3fLL);
  sort(p + 1, p + n + 1, cmp);
  for (int i = 1; i <= 10; ++i) {
    Point now = Point(0LL, 0LL);
    for (int i = 1; i <= n; ++i) {
      if (now + p[i] < now - p[i]) {
        now = now + p[i];
        ans[p[i].id] = 1;
      } else {
        now = now - p[i];
        ans[p[i].id] = -1;
      }
    }
    if (now < lst)
      for (int i = 1; i <= n; ++i) anss[i] = ans[i];
    random_shuffle(p + 1, p + n + 1);
  }
  for (int i = 1; i <= n; ++i) printf("%d ", anss[i]);
  return 0;
}