Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n;
long long nx, ny;
long long xx, yy;
long long nx1, ny1, nx2, ny2;
int main() {
  xx = 0;
  yy = 0;
  n = read();
  int ans[100005];
  long long x[100005];
  long long y[100005];
  for (long long i = 1; i <= n; i++) {
    x[i] = read();
    y[i] = read();
  }
  for (long long i = n; i > 0; i--) {
    nx = x[i];
    ny = y[i];
    nx1 = xx + nx;
    nx2 = xx - nx;
    ny1 = yy + ny;
    ny2 = yy - ny;
    if (nx1 * nx1 + ny1 * ny1 <= nx2 * nx2 + ny2 * ny2) {
      xx = nx1;
      yy = ny1;
      ans[i] = 1;
    } else {
      xx = nx2;
      yy = ny2;
      ans[i] = -1;
    }
  }
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}