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;
long long poww(long long x, long long y) {
  long long res = 1;
  x %= 1000000007;
  for (; y; y >>= 1) {
    if (y & 1) res = res * x % 1000000007;
    x = x * x % 1000000007;
  }
  return res;
}
long long val(long long x, long long y) { return x * x + y * y; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, curx, cury, i, x1, x2, temp1, temp2, ans[100005];
  cin >> n;
  multimap<pair<long long, long long>, long long> m;
  multimap<pair<long long, long long>, long long>::iterator it;
  long long x[n], y[n];
  for (i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
    if (i > 0) m.insert(make_pair(make_pair(x[i], y[i]), i));
  }
  cout << 1 << " ";
  curx = x[0];
  cury = y[0];
  memset(ans, 0, sizeof(ans));
  while (m.size() != 0) {
    it = m.lower_bound(make_pair(curx, cury));
    if (it == m.end()) it--;
    temp1 = it->first.first;
    temp2 = it->first.second;
    x1 = val(curx + temp1, cury + temp2);
    x2 = val(curx - temp1, cury - temp2);
    if (x1 > x2) {
      ans[it->second] = -1;
      m.erase(it);
      curx = curx - temp1;
      cury = cury - temp2;
    } else {
      ans[it->second] = 1;
      m.erase(it);
      curx = curx + temp1;
      cury = cury + temp2;
    }
  }
  for (i = 1; i < n; i++) cout << ans[i] << " ";
  return 0;
}