#include <bits/stdc++.h>
using namespace std;
long long norm(long long a, long long b) { return a * a + b * b; }
long double args(long long a, long long b) {
  if (b < 0) {
    a *= -1;
    b *= -1;
  }
  if (a == 0 && b == 0) return 0;
  long double F = 1.0L * a / sqrtl(1.0L * (a * a + b * b));
  long double G = acos(F);
  return G * 180.0L / 3.14159265358979323L;
}
struct State {
  long double rad;
  long long ex, ey;
  vector<pair<int, int>> A;
};
bool operator<(const State &a1, const State &a2) {
  if (a1.rad < a2.rad) return true;
  return false;
}
long long n, x[100009], y[100009], c[100009], B = 1000000000000LL;
vector<State> T;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x[i] >> y[i];
    T.push_back(State{args(x[i], y[i]), x[i], y[i],
                      vector<pair<int, int>>{make_pair(i, 1)}});
  }
  while (true) {
    sort(T.begin(), T.end());
    vector<State> U;
    for (int i = 0; i < (int)T.size() - 1; i += 2) {
      long long ax = T[i].ex, ay = T[i].ey;
      long long bx = T[i + 1].ex, by = T[i + 1].ey;
      if (norm(ax + bx, ay + by) <= B) {
        State G;
        for (int j = 0; j < T[i].A.size(); j++) G.A.push_back(T[i].A[j]);
        for (int j = 0; j < T[i + 1].A.size(); j++)
          G.A.push_back(T[i + 1].A[j]);
        G.ex = ax + bx;
        G.ey = ay + by;
        G.rad = args(G.ex, G.ey);
        U.push_back(G);
      } else if (norm(ax - bx, ay - by) <= B) {
        State G;
        for (int j = 0; j < T[i].A.size(); j++) G.A.push_back(T[i].A[j]);
        for (int j = 0; j < T[i + 1].A.size(); j++)
          G.A.push_back(
              make_pair(T[i + 1].A[j].first, T[i + 1].A[j].second * -1));
        G.ex = ax - bx;
        G.ey = ay - by;
        G.rad = args(G.ex, G.ey);
        U.push_back(G);
      } else {
        U.push_back(T[i]);
        U.push_back(T[i + 1]);
      }
    }
    if (T.size() % 2 == 1) U.push_back(T[T.size() - 1]);
    if (T.size() == U.size()) break;
    T = U;
  }
  State FinalAnswer;
  FinalAnswer.ex = (1 << 30);
  FinalAnswer.ey = (1 << 30);
  for (int i = 0; i < (1 << T.size()); i++) {
    State G;
    G.ex = 0;
    G.ey = 0;
    G.A = vector<pair<int, int>>{};
    for (int j = 0; j < T.size(); j++) {
      if ((i / (1 << j)) % 2 == 0) {
        G.ex += T[j].ex;
        G.ey += T[j].ey;
        for (int k = 0; k < T[j].A.size(); k++) G.A.push_back(T[j].A[k]);
      } else {
        G.ex -= T[j].ex;
        G.ey -= T[j].ey;
        for (int k = 0; k < T[j].A.size(); k++)
          G.A.push_back(make_pair(T[j].A[k].first, T[j].A[k].second * -1));
      }
    }
    if (norm(FinalAnswer.ex, FinalAnswer.ey) > norm(G.ex, G.ey)) {
      FinalAnswer = G;
    }
  }
  for (int i = 0; i < FinalAnswer.A.size(); i++)
    c[FinalAnswer.A[i].first] = FinalAnswer.A[i].second;
  for (int i = 1; i <= n; i++) {
    if (i >= 2) cout << " ";
    cout << c[i];
  }
  cout << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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;
long long norm(long long a, long long b) { return a * a + b * b; }
long double args(long long a, long long b) {
  if (b < 0) {
    a *= -1;
    b *= -1;
  }
  if (a == 0 && b == 0) return 0;
  long double F = 1.0L * a / sqrtl(1.0L * (a * a + b * b));
  long double G = acos(F);
  return G * 180.0L / 3.14159265358979323L;
}
struct State {
  long double rad;
  long long ex, ey;
  vector<pair<int, int>> A;
};
bool operator<(const State &a1, const State &a2) {
  if (a1.rad < a2.rad) return true;
  return false;
}
long long n, x[100009], y[100009], c[100009], B = 1000000000000LL;
vector<State> T;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x[i] >> y[i];
    T.push_back(State{args(x[i], y[i]), x[i], y[i],
                      vector<pair<int, int>>{make_pair(i, 1)}});
  }
  while (true) {
    sort(T.begin(), T.end());
    vector<State> U;
    for (int i = 0; i < (int)T.size() - 1; i += 2) {
      long long ax = T[i].ex, ay = T[i].ey;
      long long bx = T[i + 1].ex, by = T[i + 1].ey;
      if (norm(ax + bx, ay + by) <= B) {
        State G;
        for (int j = 0; j < T[i].A.size(); j++) G.A.push_back(T[i].A[j]);
        for (int j = 0; j < T[i + 1].A.size(); j++)
          G.A.push_back(T[i + 1].A[j]);
        G.ex = ax + bx;
        G.ey = ay + by;
        G.rad = args(G.ex, G.ey);
        U.push_back(G);
      } else if (norm(ax - bx, ay - by) <= B) {
        State G;
        for (int j = 0; j < T[i].A.size(); j++) G.A.push_back(T[i].A[j]);
        for (int j = 0; j < T[i + 1].A.size(); j++)
          G.A.push_back(
              make_pair(T[i + 1].A[j].first, T[i + 1].A[j].second * -1));
        G.ex = ax - bx;
        G.ey = ay - by;
        G.rad = args(G.ex, G.ey);
        U.push_back(G);
      } else {
        U.push_back(T[i]);
        U.push_back(T[i + 1]);
      }
    }
    if (T.size() % 2 == 1) U.push_back(T[T.size() - 1]);
    if (T.size() == U.size()) break;
    T = U;
  }
  State FinalAnswer;
  FinalAnswer.ex = (1 << 30);
  FinalAnswer.ey = (1 << 30);
  for (int i = 0; i < (1 << T.size()); i++) {
    State G;
    G.ex = 0;
    G.ey = 0;
    G.A = vector<pair<int, int>>{};
    for (int j = 0; j < T.size(); j++) {
      if ((i / (1 << j)) % 2 == 0) {
        G.ex += T[j].ex;
        G.ey += T[j].ey;
        for (int k = 0; k < T[j].A.size(); k++) G.A.push_back(T[j].A[k]);
      } else {
        G.ex -= T[j].ex;
        G.ey -= T[j].ey;
        for (int k = 0; k < T[j].A.size(); k++)
          G.A.push_back(make_pair(T[j].A[k].first, T[j].A[k].second * -1));
      }
    }
    if (norm(FinalAnswer.ex, FinalAnswer.ey) > norm(G.ex, G.ey)) {
      FinalAnswer = G;
    }
  }
  for (int i = 0; i < FinalAnswer.A.size(); i++)
    c[FinalAnswer.A[i].first] = FinalAnswer.A[i].second;
  for (int i = 1; i <= n; i++) {
    if (i >= 2) cout << " ";
    cout << c[i];
  }
  cout << endl;
  return 0;
}

```