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 double PI = acos(-1);
const int LIM = 15e5;
struct Vector {
  int x, y;
  Vector operator+(Vector v) { return {x + v.x, y + v.y}; }
  Vector operator-(Vector v) { return {x - v.x, y - v.y}; }
  long long operator%(Vector v) { return 1LL * x * v.y - 1LL * v.x * y; }
  long long operator*(Vector v) { return 1LL * x * v.x + 1LL * y * v.y; }
  friend double angle(Vector va, Vector vb) {
    return abs(atan2(va % vb, va * vb) * 180 / PI);
  }
  double len() { return hypot(x, y); }
  Vector operator*(int k) { return {k * x, k * y}; }
};
const int N = 1e5 + 1;
int n, sig[N];
Vector a[N];
vector<int> s[N];
void combine(int p) {
  for (int i = p; i <= p + 2; ++i)
    for (int j = i + 1; j <= p + 2; ++j) {
      if (angle(a[i], a[j]) >= 120) {
        int r = 3 * p + 3 - i - j;
        if (s[i].size() > s[j].size()) s[i].swap(s[j]);
        s[j].insert(s[j].end(), s[i].begin(), s[i].end());
        a[j] = a[i] + a[j];
        if (r == p) {
          s[i].swap(s[r]);
          a[i] = a[r];
        }
        return;
      } else if (angle(a[i], a[j]) <= 60) {
        int r = 3 * p + 3 - i - j;
        for (int id : s[j]) sig[id] *= -1;
        if (s[i].size() > s[j].size()) s[i].swap(s[j]);
        s[j].insert(s[j].end(), s[i].begin(), s[i].end());
        a[j] = a[i] - a[j];
        if (r == p) {
          s[i].swap(s[r]);
          a[i] = a[r];
        }
        return;
      }
    }
}
bool toolong(Vector v) { return v.len() > LIM; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].x >> a[i].y;
    s[i].push_back(i);
    sig[i] = 1;
  }
  if (n == 1) return cout << 1, 0;
  for (int i = 1; i <= n - 2; ++i) {
    combine(i);
  }
  if (angle(a[n - 1], a[n]) <= 90) {
    for (int id : s[n]) sig[id] *= -1;
  }
  for (int i = 1; i <= n; ++i) {
    cout << sig[i] << ' ';
  }
  return 0;
}