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;
long long x[100005];
long long y[100005];
int ans[100005];
long long sumx, sumy;
long long nowx, nowy;
long long n;
long long z;
long long zz;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  srand(time(NULL));
  for (int i = 1; i <= n; i++) {
    cin >> x[i];
    cin >> y[i];
  }
  nowx = 1000000000;
  nowy = 1000000000;
  while ((sqrt(nowx * nowx + nowy * nowy)) > 1500000) {
    nowx = 0;
    nowy = 0;
    for (int i = 1; i <= n; i++) {
      if ((nowx + x[i]) * (nowx + x[i]) + (nowy + y[i]) * (nowy + y[i]) <=
          (nowx - x[i]) * (nowx - x[i]) + (nowy - y[i]) * (nowy - y[i])) {
        z = rand() % 1000000000000000000;
        zz = rand() % 10;
        if (z == 1 && zz == 1) {
          ans[i] = -1;
          nowx -= x[i];
          nowy -= y[i];
        } else {
          ans[i] = 1;
          nowx += x[i];
          nowy += y[i];
        }
      } else {
        z = rand() % 1000000000000000000;
        zz = rand() % 10;
        if (z == 1 && zz == 1) {
          ans[i] = 1;
          nowx += x[i];
          nowy += y[i];
        } else {
          ans[i] = -1;
          nowx -= x[i];
          nowy -= y[i];
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
}
