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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const long long lim = (long long)(1e+6 * 1.5);
pair<long long, long long> operator+(const pair<long long, long long> &a,
                                     const pair<long long, long long> &b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
long long len(const pair<long long, long long> &a) {
  return a.first * a.first + a.second * a.second;
}
int main() {
  srand(1373737);
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<pair<long long, long long> > arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i].first >> arr[i].second;
  vector<int> ord(n);
  for (int i = 0; i < n; i++) ord[i] = i;
  while (true) {
    random_shuffle(ord.begin(), ord.end());
    pair<long long, long long> now = arr[ord[0]];
    vector<bool> ans(n, true);
    for (int i = 1; i < n; i++) {
      pair<long long, long long> to1 = arr[ord[i]],
                                 to2 = make_pair(-arr[ord[i]].first,
                                                 -arr[ord[i]].second);
      if (len(now + to1) < len(now + to2)) {
        ans[ord[i]] = true;
        now = now + to1;
      } else {
        ans[ord[i]] = false;
        now = now + to2;
      }
    }
    if (len(now) > lim * lim) continue;
    for (int i = 0; i < n; i++) {
      if (!ans[i])
        cout << "-1 ";
      else
        cout << "1 ";
    }
    return 0;
  }
  return 0;
}
