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 far(long long x, long long y) { return x * x + y * y; }
vector<long long> val;
vector<long long> dsu;
vector<long long> rang;
long long pred(long long a) {
  if (a == dsu[a]) {
    return a;
  } else {
    long long to = pred(dsu[a]);
    if (to != dsu[a]) {
      val[a] *= val[dsu[a]];
    }
    dsu[a] = to;
    return to;
  }
}
void unite(long long a, long long b) {
  a = pred(a);
  b = pred(b);
  if (a != b) {
    if (rang[a] < rang[b]) {
      swap(a, b);
    }
    if (val[a] == -1) {
      val[b] *= -1;
    }
    dsu[b] = a;
    if (rang[a] == rang[b]) {
      rang[a]++;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  val.resize(n);
  dsu.resize(n);
  rang.resize(n);
  vector<pair<long long, long long> > x(n);
  vector<long long> type(n, 1);
  long long stx = 0, sty = 0;
  for (long long i = 0; i < n; i++) {
    cin >> x[i].first >> x[i].second;
  }
  long long p = 1e6;
  p = p * p;
  vector<long long> ind(n);
  for (long long i = 0; i < n; i++) {
    ind[i] = i;
    dsu[i] = i;
    rang[i] = 1;
    val[i] = 1;
  }
  for (long long i = 0; i < n - 2; i++) {
    for (long long j = i + 0; j < min(n, 3 + i); j++) {
      for (long long d = j + 1; d < min(n, 3 + i); d++) {
        if (far(x[j].first + x[d].first, x[j].second + x[d].second) <= p) {
          unite(ind[j], ind[d]);
          x[d].first += x[j].first;
          x[d].second += x[j].second;
          if (j != i) {
            x[i + 1] = x[i];
            ind[i + 1] = ind[i];
          }
          j = 3 + i;
          break;
        } else if (far(x[j].first - x[d].first, x[j].second - x[d].second) <=
                   p) {
          val[pred(ind[d])] *= -1;
          unite(ind[j], ind[d]);
          x[d].first = x[j].first - x[d].first;
          x[d].second = x[j].second - x[d].second;
          if (j != i) {
            x[i + 1] = x[i];
            ind[i + 1] = ind[i];
          }
          j = 3 + i;
          break;
        }
      }
    }
  }
  long long p1 = 1.5e6;
  p1 = p1 * p1;
  if (n > 1) {
    if (far(x[n - 2].first + x[n - 1].first,
            x[n - 2].second + x[n - 1].second) <= p1) {
      unite(ind[n - 2], ind[n - 1]);
    } else {
      val[pred(ind[n - 1])] *= -1;
      unite(ind[n - 2], ind[n - 1]);
    }
  }
  for (long long i = 0; i < n; i++) {
    long long t = pred(i);
    if (t != i) {
      cout << val[t] * val[i] << " ";
    } else {
      cout << val[i] << " ";
    }
  }
  return 0;
}