#include <bits/stdc++.h>
using namespace std;
const long long MX = 1ll * 9 * 1e12 / 4;
bool comp(pair<pair<long long, long long>, int> a,
          pair<pair<long long, long long>, int> b) {
  return a.first.first * a.first.first + a.first.second * a.first.second >
         b.first.first * b.first.first + b.first.second * b.first.second;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n;
  cin >> n;
  vector<pair<pair<long long, long long>, int> > data(n);
  pair<long long, long long> v;
  for (int i = 0; i < n; ++i) {
    cin >> data[i].first.first >> data[i].first.second;
    data[i].second = i;
    v.first += data[i].first.first;
    v.second += data[i].first.second;
  }
  pair<long long, long long> cpv = v;
  vector<int> ans(n);
  for (int k = 0; k < 100000; ++k) {
    int i = 0;
    shuffle(data.begin(), data.end(), rng);
    fill(ans.begin(), ans.end(), 1);
    for (; i < n && (abs(v.first) > 1e9 || abs(v.second) > 1e9 ||
                     v.first * v.first + v.second * v.second > MX);
         ++i) {
      v.first -= 2 * data[i].first.first;
      v.second -= 2 * data[i].first.second;
      ans[data[i].second] = -1;
    }
    if (v.first * v.first + v.second * v.second <= MX) {
      for (auto x : ans) cout << x << ' ';
      cout << '\n';
      return 0;
    }
    v = cpv;
  }
  return 0;
}

### Prompt

Your task is to create 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 MX = 1ll * 9 * 1e12 / 4;
bool comp(pair<pair<long long, long long>, int> a,
          pair<pair<long long, long long>, int> b) {
  return a.first.first * a.first.first + a.first.second * a.first.second >
         b.first.first * b.first.first + b.first.second * b.first.second;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n;
  cin >> n;
  vector<pair<pair<long long, long long>, int> > data(n);
  pair<long long, long long> v;
  for (int i = 0; i < n; ++i) {
    cin >> data[i].first.first >> data[i].first.second;
    data[i].second = i;
    v.first += data[i].first.first;
    v.second += data[i].first.second;
  }
  pair<long long, long long> cpv = v;
  vector<int> ans(n);
  for (int k = 0; k < 100000; ++k) {
    int i = 0;
    shuffle(data.begin(), data.end(), rng);
    fill(ans.begin(), ans.end(), 1);
    for (; i < n && (abs(v.first) > 1e9 || abs(v.second) > 1e9 ||
                     v.first * v.first + v.second * v.second > MX);
         ++i) {
      v.first -= 2 * data[i].first.first;
      v.second -= 2 * data[i].first.second;
      ans[data[i].second] = -1;
    }
    if (v.first * v.first + v.second * v.second <= MX) {
      for (auto x : ans) cout << x << ' ';
      cout << '\n';
      return 0;
    }
    v = cpv;
  }
  return 0;
}

```