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;
const int MAX = 1E6 + 10;
int n;
int x[MAX], y[MAX];
long long IsLess(long long x, long long y, long long up) {
  return ((x) * (x)) + ((y) * (y)) <= ((up) * (up));
}
struct Part {
  long long x, y;
  vector<pair<int, int> > is;
  void MergeIn(Part& p, bool op) {
    if (((int)(is).size()) < ((int)(p.is).size())) Swap(p);
    if (op) {
      for (auto& e : p.is) e.second *= -1;
      p.x *= -1;
      p.y *= -1;
    }
    x += p.x;
    y += p.y;
    for (const auto& e : p.is) is.push_back(e);
  }
  void Swap(Part& w) {
    swap(x, w.x);
    swap(y, w.y);
    is.swap(w.is);
  }
};
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d%d", &x[i], &y[i]);
  vector<Part> nal;
  for (int i = 0; i < n; ++i) {
    Part p;
    p.x = x[i];
    p.y = y[i];
    p.is.push_back(make_pair(i, +1));
    nal.push_back(p);
  }
  while (((int)(nal).size()) > 2) {
    bool done = false;
    for (int i = 0; i < 3; ++i)
      for (int j = i + 1; j < 3; ++j) {
        if (done) break;
        if (IsLess(nal[i].x + nal[j].x, nal[i].y + nal[j].y, 1000000)) {
          nal[i].MergeIn(nal[j], false);
          nal[j].Swap(nal.back());
          nal.pop_back();
          done = true;
          break;
        }
        if (IsLess(nal[i].x - nal[j].x, nal[i].y - nal[j].y, 1000000)) {
          nal[i].MergeIn(nal[j], true);
          nal[j].Swap(nal.back());
          nal.pop_back();
          done = true;
          break;
        }
      }
  }
  if (((int)(nal).size()) == 2) {
    if (IsLess(nal[0].x + nal[1].x, nal[0].y + nal[1].y, 1500000)) {
      nal[0].MergeIn(nal[1], false);
    } else {
      nal[0].MergeIn(nal[1], true);
    }
  }
  vector<int> res(n);
  for (const auto& e : nal[0].is) res[e.first] = e.second;
  for (int i = 0; i < n; ++i) printf("%d%c", res[i], i + 1 == n ? '\n' : ' ');
  return 0;
}
