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;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
const long long N = 1000 * 1000 + 7;
long long test;
long long n;
vector<pair<pair<long long, long long>, long long>> a, b;
long long res[N];
vector<pair<pair<long long, long long>, long long>> g, G;
void ad(pair<long long, long long> &a, pair<long long, long long> b) {
  a.first += b.first;
  a.second += b.second;
}
void rem(pair<long long, long long> &a, pair<long long, long long> b) {
  a.first -= b.first;
  a.second -= b.second;
}
long long len(pair<long long, long long> a) {
  return sqrt(a.first * a.first + a.second * a.second);
}
void solve() {
  cin >> n;
  for (long long i = 1; i <= (n); ++i) {
    long long x, y;
    cin >> x >> y;
    long long id = i;
    if (x < 0) {
      id *= -1;
      x *= -1;
      y *= -1;
    }
    g.emplace_back(make_pair(x, y), id);
  }
  sort(g.begin(), g.end());
  pair<long long, long long> a, b;
  a = b = make_pair(0, 0);
  for (long long i = 0; i < (n); ++i) {
    if (i % 2) {
      g[i].first.first *= -1;
      g[i].first.second *= -1;
      g[i].second *= -1;
    }
    a.first += g[i].first.first;
    a.second += g[i].first.second;
  }
  for (long long i = 0; i < (n); ++i) {
    b.first += g[i].first.first;
    b.second += g[i].first.second;
    if (abs(a.first - 2 * b.first) <= 1000 * 1000 &&
        abs(a.second - 2 * b.second) <= 1000 * 1000) {
      for (long long j = 0; j <= (i); ++j) {
        g[j].first.first *= -1;
        g[j].first.second *= -1;
        g[j].second *= -1;
      }
      break;
    }
  }
  for (auto it : g) {
    res[abs(it.second)] = it.second;
  }
  for (long long i = 1; i <= (n); ++i) {
    cout << (res[i] > 0 ? 1 : -1) << " ";
  }
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  cout << fixed;
  test = 1;
  for (long long i = 1; i <= test; i++) {
    solve();
  }
  return 0;
}
