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;
template <class T>
void splitstr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
static void redirect(int argc, const char **argv) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (argc > 1) {
    static filebuf f;
    f.open(argv[1], ios::in);
    cin.rdbuf(&f);
    if (!cin) {
      cerr << "Failed to open '" << argv[1] << "'" << endl;
      exit(1);
    }
  }
  if (argc > 2) {
    static filebuf f;
    f.open(argv[2], ios::out | ios::trunc);
    cout.rdbuf(&f);
    if (!cout) {
      cerr << "Failed to open '" << argv[2] << "'" << endl;
    }
  }
}
static inline long long dot(const std::complex<long long> &a,
                            const std::complex<long long> &b) {
  return real(conj(a) * b);
}
static inline long long cross(const std::complex<long long> &a,
                              const std::complex<long long> &b) {
  return imag(conj(a) * b);
}
static inline long long cross(const std::complex<long long> &a,
                              const std::complex<long long> &b,
                              const std::complex<long long> &c) {
  return cross(b - a, c - a);
}
struct lpnt {
  std::complex<long long> pos;
  int idx;
};
class CompareAngle {
 public:
  bool operator()(const lpnt &a, const lpnt &b) const {
    long long c = cross(a.pos, b.pos);
    if (c != 0)
      return c > 0;
    else
      return a.idx < b.idx;
  }
};
static bool below(const std::complex<long long> &p) {
  return p.imag() < 0 || (p.imag() == 0 && p.real() < 0);
}
int main(int argc, const char **argv) {
  redirect(argc, argv);
  int N;
  cin >> N;
  set<lpnt, CompareAngle> pnts;
  vector<std::complex<long long> > orig(N);
  vector<unsigned char> flip(N);
  vector<int> parent(N, -1);
  for (int i = 0; i < N; i++) {
    int x, y;
    cin >> x >> y;
    std::complex<long long> p(x, y);
    orig[i] = p;
    if (below(p)) {
      p = -p;
      flip[i] = true;
    }
    pnts.insert(lpnt{p, i});
  }
  while (((long long)(pnts).size()) > 1) {
    for (auto u = pnts.begin(); u != pnts.end(); ++u) {
      auto v = next(u);
      if (v == pnts.end()) v = pnts.begin();
      long long du = dot(u->pos, u->pos);
      long long dv = dot(v->pos, v->pos);
      for (int i = 0; i < 2; i++)
        for (int j = 0; j < 2; j++) {
          std::complex<long long> sum =
              (i ? -u->pos : u->pos) + (j ? -v->pos : v->pos);
          long long ds = dot(sum, sum);
          if (!below(sum) && (ds <= du || ds <= dv)) {
            int idx = ((long long)(parent).size());
            flip[u->idx] ^= i;
            flip[v->idx] ^= j;
            parent[u->idx] = idx;
            parent[v->idx] = idx;
            pnts.erase(u);
            pnts.erase(v);
            pnts.insert(lpnt{sum, idx});
            parent.push_back(-1);
            flip.push_back(false);
            goto have_found;
          }
        }
    }
    break;
  have_found:;
  }
  if (((long long)(pnts).size()) > 1) {
    assert(((long long)(pnts).size()) == 2);
    auto u = pnts.begin();
    auto v = next(u);
    if (dot(u->pos, v->pos) > 0) flip[v->idx] ^= true;
  }
  for (int i = ((long long)(parent).size()) - 1; i >= 0; i--)
    if (parent[i] != -1) flip[i] ^= flip[parent[i]];
  for (int i = 0; i < N; i++) cout << (flip[i] ? -1 : 1) << ' ';
  cout << '\n';
  return 0;
}