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;
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
int main() {
  int n, x, y;
  cin >> n;
  vector<vector<int> > v;
  vector<int> g;
  for (int i = 0; i < n; i++) {
    g.clear();
    cin >> x >> y;
    g.push_back(x);
    g.push_back(y);
    g.push_back(i);
    v.push_back(g);
  }
  int ans[n];
  for (int l = 1; l <= 10; l++) {
    random_shuffle(v.begin(), v.end());
    long long p = 0;
    long long a, b;
    a = b = 0;
    for (int i = 0; i < n; i++) {
      if ((a + v[i][0]) * (a + v[i][0]) + (b + v[i][1]) * (b + v[i][1]) >
          (a - v[i][0]) * (a - v[i][0]) + (b - v[i][1]) * (b - v[i][1])) {
        ans[v[i][2]] = -1;
        a = a - v[i][0];
        b = b - v[i][1];
        p = (a) * (a) + (b) * (b);
      } else {
        ans[v[i][2]] = 1;
        a += v[i][0];
        b += v[i][1];
        p = (a) * (a) + (b) * (b);
      }
    }
    if (p <= 2250000000000LL) {
      break;
    }
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << '\n';
  return 0;
}
