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 
#include <bits/stdc++.h>
using namespace std;
string str[4];
long long x[300005], y[300005], dir[300005], vis[300005];
vector<long long> v;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, k, p, a, b, c, a1, b1, a2, b2;
  srand(time(NULL));
  while (cin >> n) {
    double lim = (1.5 * 1000000);
    lim = lim * lim;
    for (long long i = 0; i < n; i++) {
      cin >> x[i] >> y[i];
      v.push_back(i);
    }
    while (1) {
      a = b = 0;
      random_shuffle(v.begin(), v.end());
      for (long long j = 0; j < v.size(); j++) {
        long long i = v[j];
        a1 = a + x[i];
        b1 = b + y[i];
        a2 = a - x[i];
        b2 = b - y[i];
        k = a1 * a1 + b1 * b1;
        m = a2 * a2 + b2 * b2;
        if (k <= m) {
          dir[i] = 1;
          a = a1;
          b = b1;
          p = k;
        } else {
          dir[i] = -1;
          a = a2;
          b = b2;
          p = m;
        }
      }
      double tmp = (double)(p);
      if (tmp <= lim) break;
    }
    for (long long i = 0; i < n; i++) cout << dir[i] << " ";
    cout << endl;
    v.clear();
  }
  return 0;
}
