#include <bits/stdc++.h>
using namespace std;
long long int n;
struct Vec {
  long long int x;
  long long int y;
  long long int ind;
  long long int mod;
};
struct Vec v[200005];
long long int ans[200005];
int comp(const void *p, const void *q) {
  struct Vec *a = (struct Vec *)p;
  struct Vec *b = (struct Vec *)q;
  if (a->mod > b->mod) {
    return 1;
  }
  return -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  long long int i;
  for (i = 1; i <= n; ++i) {
    cin >> v[i].x >> v[i].y;
    v[i].ind = i;
    v[i].mod = v[i].x * v[i].x + v[i].y * v[i].y;
  }
  long double curx = 0;
  long double cury = 0;
  long long int p = 1500000;
  p = p * p;
  for (i = n; i >= 1; --i) {
    if ((curx + v[i].x) * (curx + v[i].x) + (cury + v[i].y) * (cury + v[i].y) <
        (curx - v[i].x) * (curx - v[i].x) + (cury - v[i].y) * (cury - v[i].y)) {
      curx += v[i].x;
      cury += v[i].y;
      ans[v[i].ind] = 1;
    } else {
      curx -= v[i].x;
      cury -= v[i].y;
      ans[v[i].ind] = -1;
    }
  }
  if (curx * curx + cury * cury <= p && curx * curx + cury * cury >= 0) {
    for (i = 1; i <= n; ++i) {
      cout << ans[i] << " ";
    }
    cout << endl;
  } else {
    cout << "-1" << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
long long int n;
struct Vec {
  long long int x;
  long long int y;
  long long int ind;
  long long int mod;
};
struct Vec v[200005];
long long int ans[200005];
int comp(const void *p, const void *q) {
  struct Vec *a = (struct Vec *)p;
  struct Vec *b = (struct Vec *)q;
  if (a->mod > b->mod) {
    return 1;
  }
  return -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  long long int i;
  for (i = 1; i <= n; ++i) {
    cin >> v[i].x >> v[i].y;
    v[i].ind = i;
    v[i].mod = v[i].x * v[i].x + v[i].y * v[i].y;
  }
  long double curx = 0;
  long double cury = 0;
  long long int p = 1500000;
  p = p * p;
  for (i = n; i >= 1; --i) {
    if ((curx + v[i].x) * (curx + v[i].x) + (cury + v[i].y) * (cury + v[i].y) <
        (curx - v[i].x) * (curx - v[i].x) + (cury - v[i].y) * (cury - v[i].y)) {
      curx += v[i].x;
      cury += v[i].y;
      ans[v[i].ind] = 1;
    } else {
      curx -= v[i].x;
      cury -= v[i].y;
      ans[v[i].ind] = -1;
    }
  }
  if (curx * curx + cury * cury <= p && curx * curx + cury * cury >= 0) {
    for (i = 1; i <= n; ++i) {
      cout << ans[i] << " ";
    }
    cout << endl;
  } else {
    cout << "-1" << endl;
  }
  return 0;
}

```