#include <bits/stdc++.h>
using namespace std;
long long dist(long long x, long long y) { return x * x + y * y; }
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<pair<int, int>, int> > vec;
  for (int i = 0, x, y; i < n; i++) {
    scanf("%d %d", &x, &y);
    vec.push_back(make_pair(make_pair(x, y), i));
  }
  while (true) {
    long long nowx = 0, nowy = 0;
    vector<int> res(n);
    for (int i = 0; i < vec.size(); i++) {
      long long x = vec[i].first.first;
      long long y = vec[i].first.second;
      long long dist1 = dist(nowx + x, nowy + y);
      long long dist2 = dist(nowx - x, nowy - y);
      if (dist1 <= dist2) {
        nowx += x;
        nowy += y;
        res[vec[i].second] = 1;
      } else {
        nowx -= x;
        nowy -= y;
        res[vec[i].second] = -1;
      }
    }
    if (nowx * nowx + nowy * nowy <= 225LL * (long long)1e10) {
      for (int i = 0; i < res.size(); i++) {
        if (i) printf(" ");
        printf("%d", res[i]);
      }
      puts("");
      return 0;
    }
    srand(time(NULL));
    for (int i = 0; i < 1000; i++) swap(vec[rand() % n], vec[rand() % 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;
long long dist(long long x, long long y) { return x * x + y * y; }
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<pair<int, int>, int> > vec;
  for (int i = 0, x, y; i < n; i++) {
    scanf("%d %d", &x, &y);
    vec.push_back(make_pair(make_pair(x, y), i));
  }
  while (true) {
    long long nowx = 0, nowy = 0;
    vector<int> res(n);
    for (int i = 0; i < vec.size(); i++) {
      long long x = vec[i].first.first;
      long long y = vec[i].first.second;
      long long dist1 = dist(nowx + x, nowy + y);
      long long dist2 = dist(nowx - x, nowy - y);
      if (dist1 <= dist2) {
        nowx += x;
        nowy += y;
        res[vec[i].second] = 1;
      } else {
        nowx -= x;
        nowy -= y;
        res[vec[i].second] = -1;
      }
    }
    if (nowx * nowx + nowy * nowy <= 225LL * (long long)1e10) {
      for (int i = 0; i < res.size(); i++) {
        if (i) printf(" ");
        printf("%d", res[i]);
      }
      puts("");
      return 0;
    }
    srand(time(NULL));
    for (int i = 0; i < 1000; i++) swap(vec[rand() % n], vec[rand() % n]);
  }
  return 0;
}

```