#include <bits/stdc++.h>
using namespace std;
const long long big = (long long)1500000 * (long long)1500000;
struct Vector {
  long long x;
  long long y;
  int w;
  long long sqNorm() { return (x * x + y * y); }
  void sub(Vector oth) {
    x -= oth.x;
    y -= oth.y;
  }
  void add(Vector oth) {
    x += oth.x;
    y += oth.y;
  }
};
Vector vec[100001];
int ans[100001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a, b;
    cin >> a >> b;
    vec[i] = {a, b, i};
  }
  while (true) {
    random_shuffle(vec + 1, vec + 1 + n);
    Vector p = {0, 0, 0};
    for (int i = 1; i <= n; i++) {
      int w = vec[i].w;
      p.sub(vec[i]);
      Vector s = p;
      p.add(vec[i]);
      p.add(vec[i]);
      Vector a = p;
      if (a.sqNorm() < s.sqNorm()) {
        p = a;
        ans[w] = 1;
      } else {
        p = s;
        ans[w] = -1;
      }
    }
    if (p.sqNorm() <= big) break;
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " \n"[i == n];
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 big = (long long)1500000 * (long long)1500000;
struct Vector {
  long long x;
  long long y;
  int w;
  long long sqNorm() { return (x * x + y * y); }
  void sub(Vector oth) {
    x -= oth.x;
    y -= oth.y;
  }
  void add(Vector oth) {
    x += oth.x;
    y += oth.y;
  }
};
Vector vec[100001];
int ans[100001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a, b;
    cin >> a >> b;
    vec[i] = {a, b, i};
  }
  while (true) {
    random_shuffle(vec + 1, vec + 1 + n);
    Vector p = {0, 0, 0};
    for (int i = 1; i <= n; i++) {
      int w = vec[i].w;
      p.sub(vec[i]);
      Vector s = p;
      p.add(vec[i]);
      p.add(vec[i]);
      Vector a = p;
      if (a.sqNorm() < s.sqNorm()) {
        p = a;
        ans[w] = 1;
      } else {
        p = s;
        ans[w] = -1;
      }
    }
    if (p.sqNorm() <= big) break;
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " \n"[i == n];
  return 0;
}

```