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;
const int N = 101010;
pair<long long, long long> operator+(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return {a.first + b.first, a.second + b.second};
}
pair<long long, long long> operator-(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return {a.first - b.first, a.second - b.second};
}
long long sqr(long long x) { return x * x; }
long long norm(const pair<long long, long long>& a) {
  return sqr(a.first) + sqr(a.second);
}
void scan(pair<long long, long long>& tp) {
  scanf("%lld%lld", &tp.first, &tp.second);
}
int n, sgn[N], ord[N];
pair<long long, long long> p[N], cur;
const long long LIM = 1500000LL * 1500000LL;
void output() {
  for (int i = 0; i < n; i++) printf("%d%c", sgn[i], " \n"[i + 1 == n]);
  exit(0);
}
void go(int ptr) {
  if (ptr == n) {
    if (norm(cur) <= LIM) output();
    return;
  }
  int i = ord[ptr];
  int si = 1;
  if (norm(cur - p[i]) < norm(cur + p[i])) si = -1;
  for (int _ = 0; _ < 2; _++, si = -si) {
    sgn[i] = si;
    if (si == 1)
      cur = cur + p[i];
    else
      cur = cur - p[i];
    go(ptr + 1);
    if (si == 1)
      cur = cur - p[i];
    else
      cur = cur + p[i];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scan(p[i]);
  for (int i = 0; i < n; i++) ord[i] = i;
  random_shuffle(ord, ord + n);
  go(0);
}