#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const long long int oo = numeric_limits<long long int>::max();
const long long int BOUND = 1.5e6;
const long long int SQ_BOUND = BOUND * BOUND;
inline long long int norm_sq(const complex<long long int> &z) {
  return (z * conj(z)).real();
}
int n;
long long int s[N];
complex<long long int> o, arr[N];
int run() {
  long long int mn = oo;
  int opt = -1;
  for (int i = 1; i <= n; ++i) {
    s[i] = -s[i];
    complex<long long int> aux = o + 2 * s[i] * arr[i];
    if (norm_sq(aux) < mn) {
      mn = norm_sq(aux);
      opt = i;
    }
    s[i] = -s[i];
  }
  return opt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x, y;
    cin >> x >> y;
    arr[i] = {x, y};
  }
  bool success;
  while (1) {
    o = {0, 0};
    for (int i = 1; i <= n; ++i) {
      s[i] = (rand() & 1) ? 1 : -1;
      o += s[i] * arr[i];
    }
    success = true;
    while (norm_sq(o) > SQ_BOUND) {
      int opt = run();
      if (!~opt) {
        success = false;
        break;
      }
      s[opt] = -s[opt];
      o += 2 * s[opt] * arr[opt];
    }
    if (success) {
      for (int j = 1; j <= n; j++) cout << s[j] << " \n"[j == n];
      break;
    }
  }
  assert(success);
  return 0;
}

### Prompt

In cpp, your task is to solve 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 int N = 2e5 + 5;
const long long int oo = numeric_limits<long long int>::max();
const long long int BOUND = 1.5e6;
const long long int SQ_BOUND = BOUND * BOUND;
inline long long int norm_sq(const complex<long long int> &z) {
  return (z * conj(z)).real();
}
int n;
long long int s[N];
complex<long long int> o, arr[N];
int run() {
  long long int mn = oo;
  int opt = -1;
  for (int i = 1; i <= n; ++i) {
    s[i] = -s[i];
    complex<long long int> aux = o + 2 * s[i] * arr[i];
    if (norm_sq(aux) < mn) {
      mn = norm_sq(aux);
      opt = i;
    }
    s[i] = -s[i];
  }
  return opt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x, y;
    cin >> x >> y;
    arr[i] = {x, y};
  }
  bool success;
  while (1) {
    o = {0, 0};
    for (int i = 1; i <= n; ++i) {
      s[i] = (rand() & 1) ? 1 : -1;
      o += s[i] * arr[i];
    }
    success = true;
    while (norm_sq(o) > SQ_BOUND) {
      int opt = run();
      if (!~opt) {
        success = false;
        break;
      }
      s[opt] = -s[opt];
      o += 2 * s[opt] * arr[opt];
    }
    if (success) {
      for (int j = 1; j <= n; j++) cout << s[j] << " \n"[j == n];
      break;
    }
  }
  assert(success);
  return 0;
}

```