#include <bits/stdc++.h>
using namespace std;
struct vec {
  int x;
  int y;
};
int n, res[100010];
vec v[100010];
inline double dist(int x, int y) { return sqrt(pow(x, 2) + pow(y, 2)); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> v[i].x >> v[i].y;
  }
  int x = v[0].x, y = v[0].y;
  res[0] = 1;
  int j1 = 1, j2 = 2;
  while (j2 < n) {
    double d[4];
    d[0] = dist(x + v[j1].x, y + v[j1].y);
    d[1] = dist(x - v[j1].x, y - v[j1].y);
    d[2] = dist(x + v[j2].x, y + v[j2].y);
    d[3] = dist(x - v[j2].x, y - v[j2].y);
    int mn = min_element(d, d + 4) - d;
    if (mn == 0) {
      res[j1] = 1;
      x += v[j1].x;
      y += v[j1].y;
      j1 = j2;
      ++j2;
    } else if (mn == 1) {
      res[j1] = -1;
      x -= v[j1].x;
      y -= v[j1].y;
      j1 = j2;
      ++j2;
    } else if (mn == 2) {
      res[j2] = 1;
      x += v[j2].x;
      y += v[j2].y;
      ++j2;
    } else {
      res[j2] = -1;
      x -= v[j2].x;
      y -= v[j2].y;
      ++j2;
    }
  }
  double d[2];
  d[0] = dist(x + v[j1].x, y + v[j1].y);
  d[1] = dist(x - v[j1].x, y - v[j1].y);
  if (d[0] <= d[1])
    res[j1] = 1;
  else
    res[j1] = -1;
  for (int i = 0; i < n; ++i) cout << res[i] << ' ';
  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;
struct vec {
  int x;
  int y;
};
int n, res[100010];
vec v[100010];
inline double dist(int x, int y) { return sqrt(pow(x, 2) + pow(y, 2)); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> v[i].x >> v[i].y;
  }
  int x = v[0].x, y = v[0].y;
  res[0] = 1;
  int j1 = 1, j2 = 2;
  while (j2 < n) {
    double d[4];
    d[0] = dist(x + v[j1].x, y + v[j1].y);
    d[1] = dist(x - v[j1].x, y - v[j1].y);
    d[2] = dist(x + v[j2].x, y + v[j2].y);
    d[3] = dist(x - v[j2].x, y - v[j2].y);
    int mn = min_element(d, d + 4) - d;
    if (mn == 0) {
      res[j1] = 1;
      x += v[j1].x;
      y += v[j1].y;
      j1 = j2;
      ++j2;
    } else if (mn == 1) {
      res[j1] = -1;
      x -= v[j1].x;
      y -= v[j1].y;
      j1 = j2;
      ++j2;
    } else if (mn == 2) {
      res[j2] = 1;
      x += v[j2].x;
      y += v[j2].y;
      ++j2;
    } else {
      res[j2] = -1;
      x -= v[j2].x;
      y -= v[j2].y;
      ++j2;
    }
  }
  double d[2];
  d[0] = dist(x + v[j1].x, y + v[j1].y);
  d[1] = dist(x - v[j1].x, y - v[j1].y);
  if (d[0] <= d[1])
    res[j1] = 1;
  else
    res[j1] = -1;
  for (int i = 0; i < n; ++i) cout << res[i] << ' ';
  return 0;
}

```