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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
pair<long long int, long long int> ar[100009];
long long int curx, cury, ax, ay, bx, by, mrk[100009];
class cmp {
 public:
  bool operator()(long long int& a, long long int& b) {
    ax = 2 * ar[a].first;
    ay = 2 * ar[a].second;
    bx = 2 * ar[b].first;
    by = 2 * ar[b].second;
    if ((ax * ax + ay * ay - 2 * (curx * ax + cury * ay)) >
        (bx * bx + by * by - 2 * (curx * bx + cury * by)))
      return true;
    else
      return false;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, i, u, v;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> u >> v;
    curx += u;
    cury += v;
    ar[i] = pair<long long int, long long int>(u, v);
  }
  priority_queue<long long int, vector<long long int>, cmp> pq;
  for (i = 1; i <= n; i++) pq.push(i);
  while (curx >= 1500000 || cury >= 1500000 ||
         (4 * (curx * curx + cury * cury) >= 9 * 1e12)) {
    u = pq.top();
    curx -= 2 * ar[u].first;
    cury -= 2 * ar[u].second;
    pq.pop();
    mrk[u] = 1;
  }
  for (i = 1; i <= n; i++) {
    if (mrk[i])
      cout << -1 << " ";
    else
      cout << 1 << " ";
  }
  cout << "\n";
  return 0;
}