#include <bits/stdc++.h>
using namespace std;
const long long q = 1500000;
pair<int, int> a[100050];
int num[100050], n;
int fact[100050];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].first >> a[i].second;
    num[i] = i;
  }
  long long ans = q * q;
  int qwe = n;
  while (1) {
    long long x = 0, y = 0;
    for (int i = 1; i <= n; ++i) {
      if ((x + a[i].first) * (x + a[i].first) +
              (y + a[i].second) * (y + a[i].second) >
          (x - a[i].first) * (x - a[i].first) +
              (y - a[i].second) * (y - a[i].second))
        x -= a[i].first, y -= a[i].second, fact[num[i]] = -1;
      else
        x += a[i].first, y += a[i].second, fact[num[i]] = 1;
    }
    if (x * x + y * y <= ans) {
      for (int i = 1; i <= n; i++) {
        cout << fact[i] << " ";
      }
      cout << "\n";
      return 0;
    }
    swap(a[2].second, a[qwe].second);
    swap(a[2].first, a[qwe].first);
    swap(num[2], num[qwe--]);
  }
  return 0;
}

### Prompt

Generate a CPP solution to the following 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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long q = 1500000;
pair<int, int> a[100050];
int num[100050], n;
int fact[100050];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].first >> a[i].second;
    num[i] = i;
  }
  long long ans = q * q;
  int qwe = n;
  while (1) {
    long long x = 0, y = 0;
    for (int i = 1; i <= n; ++i) {
      if ((x + a[i].first) * (x + a[i].first) +
              (y + a[i].second) * (y + a[i].second) >
          (x - a[i].first) * (x - a[i].first) +
              (y - a[i].second) * (y - a[i].second))
        x -= a[i].first, y -= a[i].second, fact[num[i]] = -1;
      else
        x += a[i].first, y += a[i].second, fact[num[i]] = 1;
    }
    if (x * x + y * y <= ans) {
      for (int i = 1; i <= n; i++) {
        cout << fact[i] << " ";
      }
      cout << "\n";
      return 0;
    }
    swap(a[2].second, a[qwe].second);
    swap(a[2].first, a[qwe].first);
    swap(num[2], num[qwe--]);
  }
  return 0;
}

```