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;
int ans[1010101];
int n;
struct node {
  double a, b;
  int id;
} x[1010101];
signed main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x[i].a >> x[i].b;
    x[i].id = i;
  }
  while (1) {
    random_shuffle(x + 1, x + 1 + n);
    double X = x[1].a, Y = x[1].b;
    for (int i = 2; i <= n; i++) {
      if ((X + x[i].a) * (X + x[i].a) + (Y + x[i].b) * (Y + x[i].b) <
          (X - x[i].a) * (X - x[i].a) + (Y - x[i].b) * (Y - x[i].b)) {
        X += x[i].a;
        Y += x[i].b;
      } else {
        X -= x[i].a;
        Y -= x[i].b;
      }
    }
    if (X * X + Y * Y > pow(1.5e6, 2)) continue;
    X = x[1].a, Y = x[1].b;
    ans[x[1].id] = 1;
    for (int i = 2; i <= n; i++) {
      if ((X + x[i].a) * (X + x[i].a) + (Y + x[i].b) * (Y + x[i].b) <
          (X - x[i].a) * (X - x[i].a) + (Y - x[i].b) * (Y - x[i].b)) {
        ans[x[i].id] = 1;
        X += x[i].a;
        Y += x[i].b;
      } else {
        ans[x[i].id] = -1;
        X -= x[i].a;
        Y -= x[i].b;
      }
    }
    for (int i = 1; i <= n; i++) {
      printf(~ans[i] ? "1 " : "-1 ");
    }
    return 0;
  }
  return 0;
}
