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;
const int max_n = 100011, log_n = 32, max_m = 111, mod = 1000000007,
          inf = 1011111111;
int n, ans[max_n];
pair<int, int> a[max_n];
long long dist(pair<int, int> &a) {
  return 1LL * a.first * a.first + 1LL * a.second * a.second;
}
long long dist(int x, int y) { return 1LL * x * x + 1LL * y * y; }
bool is_good(pair<int, int> &a, pair<int, int> &b) {
  return sqrt(dist(a)) * sqrt(dist(b)) <=
         abs(2LL * a.first * b.first + 2LL * a.second * b.second);
}
int coeff(pair<int, int> a, pair<int, int> b) {
  if (dist(a.first + b.first, a.second + b.second) <
      dist(a.first - b.first, a.second - b.second))
    return 1;
  return -1;
}
pair<int, int> sum(pair<int, int> &a, pair<int, int> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
pair<int, int> mul(pair<int, int> &a, int b) {
  return make_pair(a.first * b, a.second * b);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &a[i].first, &a[i].second);
  }
  if (n < 2) {
    for (int i = 0; i < n; ++i) cout << "1 ";
    return 0;
  }
  pair<int, int> x = a[0];
  pair<pair<int, int>, vector<int> > y = make_pair(a[1], vector<int>(1, 1));
  ans[0] = ans[1] = 1;
  for (int i = 2; i < n; ++i) {
    if (is_good(x, y.first)) {
      int c = coeff(x, y.first);
      if (c == -1) {
        for (int q = 0; q < y.second.size(); ++q) {
          ans[y.second[q]] *= -1;
        }
      }
      x = sum(x, mul(y.first, c));
      y = make_pair(a[i], vector<int>(1, i));
      ans[i] = 1;
    } else {
      if (is_good(x, a[i])) {
        int c = coeff(x, a[i]);
        ans[i] = c;
        x = sum(x, mul(a[i], c));
      } else {
        assert(is_good(y.first, a[i]));
        int c = coeff(y.first, a[i]);
        ans[i] = c;
        y.first = sum(y.first, mul(a[i], c));
        y.second.push_back(i);
      }
    }
    assert(dist(x) <= 1000000000000LL);
    assert(dist(y.first) <= 1000000000000LL);
  }
  int c = coeff(x, y.first);
  if (c == -1) {
    for (int q = 0; q < y.second.size(); ++q) {
      ans[y.second[q]] *= -1;
    }
  }
  x = sum(x, mul(y.first, c));
  assert(dist(x) <= 2250000000000LL);
  for (int i = 0; i < n; ++i) {
    printf("%d ", ans[i]);
  }
  return 0;
}