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>
using namespace std;
const double PI = 2.0 * acos(0.0);
double EPSILON = 1e-9;
double INFTY = 1e200;
struct vector2 {
  double x, y;
  int idx;
  explicit vector2(double x_ = 0, double y_ = 0) : x(x_), y(y_) {}
  bool operator==(const vector2& rhs) const { return x == rhs.x && y == rhs.y; }
  bool operator<(const vector2& rhs) const {
    return x != rhs.x ? x < rhs.x : y < rhs.y;
  }
  vector2 operator+(const vector2& rhs) const {
    return vector2(x + rhs.x, y + rhs.y);
  }
  vector2 operator-(const vector2& rhs) const {
    return vector2(x - rhs.x, y - rhs.y);
  }
  vector2 operator*(double rhs) const { return vector2(x * rhs, y * rhs); }
  double norm() const { return hypot(x, y); }
  vector2 normalize() const { return vector2(x / norm(), y / norm()); }
  double polar() const { return fmod(atan2(y, x) + 2 * PI, 2 * PI); }
  double dot(const vector2& rhs) const { return x * rhs.x + y * rhs.y; }
  double cross(const vector2& rhs) const { return x * rhs.y - rhs.x * y; }
  vector2 project(const vector2& rhs) const {
    vector2 r = rhs.normalize();
    return r * r.dot(*this);
  }
} p[100004];
bool cmp(vector2 v, vector2 u) { return v.norm() > u.norm(); }
int n;
int main() {
  std::srand(unsigned(std::time(0)));
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i].x >> p[i].y;
    p[i].idx = i;
  }
  int ans[100004];
  while (1) {
    random_shuffle(p, p + n);
    vector2 v = p[0];
    ans[p[0].idx] = 1;
    for (int i = 1; i < n; i++) {
      vector2 u1 = v + p[i];
      vector2 u2 = v - p[i];
      if (u1.norm() < u2.norm()) {
        v = u1;
        ans[p[i].idx] = 1;
      } else {
        v = u2;
        ans[p[i].idx] = -1;
      }
    }
    if (v.norm() <= 1.5e6) break;
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  return 0;
}
