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 n;
vector<pair<long long, long long> > vec;
vector<long long> mvs;
bool done;
void printPath() {
  for (long long i = 0; i < n; i++) printf("%d ", mvs[i]);
}
double dist(long long x, long long y) { return sqrt(x * x + y * y); }
void greed(long long i, long long x, long long y) {
  if (done) return;
  if (i == n) {
    if (dist(x, y) <= 1.5 * 1e6) {
      done = true;
      printPath();
    }
    return;
  }
  double p = dist(x + vec[i].first, y + vec[i].second),
         nn = dist(x - vec[i].first, y - vec[i].second);
  if (p < nn) {
    mvs[i] = 1;
    greed(i + 1, x + vec[i].first, y + vec[i].second);
    mvs[i] = -1;
    greed(i + 1, x - vec[i].first, y - vec[i].second);
  } else {
    mvs[i] = -1;
    greed(i + 1, x - vec[i].first, y - vec[i].second);
    mvs[i] = 1;
    greed(i + 1, x + vec[i].first, y + vec[i].second);
  }
}
signed main() {
  done = false;
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  vec.resize(n + 2);
  mvs.resize(n + 2);
  for (long long i = 0; i < n; i++) cin >> vec[i].first >> vec[i].second;
  greed(0, 0, 0);
  return 0;
}