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 point {
  long long x;
  long long y;
  int p;
};
const long long UP = 2250000000000;
const int MAXN = 1e5 + 100;
int n;
point c[MAXN];
long long visit[MAXN];
long long sqr(long long x) { return x * x; }
bool splay() {
  random_shuffle(c + 1, c + n + 1);
  long long nx = 0, ny = 0;
  for (int i = 1; i < n + 1; ++i) {
    long long t1 = sqr(nx + c[i].x) + sqr(ny + c[i].y),
              t2 = sqr(nx - c[i].x) + sqr(ny - c[i].y);
    if (t1 < t2) {
      visit[c[i].p] = 1;
      nx += c[i].x;
      ny += c[i].y;
    } else {
      visit[c[i].p] = -1;
      nx -= c[i].x;
      ny -= c[i].y;
    }
  }
  return sqr(nx) + sqr(ny) <= UP;
}
int main() {
  srand((unsigned long long)(new char));
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n + 1; ++i) {
    cin >> c[i].x >> c[i].y;
    c[i].p = i;
  }
  while (!splay())
    ;
  for (int i = 1; i < n + 1; ++i) cout << visit[i] << " ";
  cout << endl;
  return 0;
}
