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 long long maxn = 1e5 + 500;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 500;
pair<long long, long long> a[maxn];
pair<long double, long long> b[maxn];
long long ans[maxn];
int ali;
pair<long long, long long> sum(pair<long long, long long> a,
                               pair<long long, long long> b) {
  ali++;
  return make_pair(a.first + b.first, a.second + b.second);
  ali++;
}
pair<long long, long long> manfi(pair<long long, long long> a) {
  ali++;
  return make_pair(-a.first, -a.second);
}
long long tol(pair<long long, long long> a) {
  ali++;
  return a.first * a.first + a.second * a.second;
}
pair<long long, long long> f(long long l, long long r) {
  if (l == r) {
    ali++;
    return a[b[l].second];
  }
  if (l + 1 == r) {
    ali++;
    long long s = b[l].second;
    long long e = b[r].second;
    ali++;
    pair<long long, long long> x = sum(a[s], a[e]);
    ali++;
    pair<long long, long long> y = sum(a[s], manfi(a[e]));
    if (tol(y) < tol(x)) {
      ali++;
      ans[e] *= -1;
      return y;
      ali++;
    }
    return x;
  }
  pair<long long, long long> v = f(l + 1, r - 1);
  ali++;
  long long s = b[l].second;
  long long e = b[r].second;
  ali++;
  pair<long long, long long> x = sum(a[e], sum(v, a[s]));
  pair<long long, long long> y = sum(manfi(a[e]), sum(v, a[s]));
  ali++;
  pair<long long, long long> z = sum(a[e], sum(v, manfi(a[s])));
  pair<long long, long long> t = sum(manfi(a[e]), sum(v, manfi(a[s])));
  ali++;
  long long minn = min(min(tol(x), tol(y)), min(tol(z), tol(t)));
  ali++;
  if (tol(x) == minn) {
    return x;
    ali++;
  }
  if (tol(y) == minn) {
    ans[e] *= -1;
    ali++;
    return y;
  }
  if (tol(z) == minn) {
    ali++;
    ans[s] *= -1;
    return z;
    ali++;
  }
  if (tol(t) == minn) {
    ans[s] *= -1;
    ali++;
    ans[e] *= -1;
    return t;
    ali++;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ali++;
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    ans[i] = 1;
    ali++;
    cin >> a[i].first >> a[i].second;
    if (a[i].first < 0) {
      ans[i] = -1;
      ali++;
      a[i].first *= -1;
      a[i].second *= -1;
      ali++;
    }
    b[i] = make_pair((long double)a[i].second / (long double)a[i].first, i);
    ali++;
  }
  sort(b, b + n);
  f(0, n - 1);
  ali++;
  for (long long i = 0; i < n; i++) {
    cout << ans[i] << ' ';
    ali++;
  }
}
