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 pii {
  int x, y, idx;
};
const int maxn = 100005;
pii A[maxn];
int ans[maxn];
int n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    A[i] = {x, y, i};
  }
  while (true) {
    random_shuffle(A, A + n);
    long long X = 0, Y = 0;
    for (int i = 0; i < n; i++) {
      long long x = A[i].x;
      long long y = A[i].y;
      long long x1 = X - x, y1 = Y - y, x2 = X + x, y2 = Y + y;
      if (x1 * x1 + y1 * y1 < x2 * x2 + y2 * y2) {
        X = x1, Y = y1;
        ans[A[i].idx] = -1;
      } else {
        X = x2, Y = y2;
        ans[A[i].idx] = 1;
      }
    }
    if (X * X + Y * Y <= 2000000000000LL) {
      for (int i = 0; i < n; i++) {
        cout << ans[i] << ' ';
      }
      return 0;
    }
  }
  return 0;
}
