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 MAXN = 3e5 + 42;
int a[MAXN];
int n;
long long dist(long long x1, long long y1) { return x1 * x1 + y1 * y1; }
int arr[MAXN];
pair<int, int> pos[MAXN];
int ans[MAXN];
const long long GG = 15e5;
const long long MAXDIST = GG * GG;
inline bool solve() {
  random_shuffle(arr, arr + n);
  long long x1 = 0, y1 = 0;
  for (int i = 0; i < n; ++i) {
    int p, q;
    p = pos[arr[i]].first;
    q = pos[arr[i]].second;
    if (dist(x1 + p, y1 + q) < dist(x1 - p, y1 - q)) {
      ans[arr[i]] = 1;
      x1 += p;
      y1 += q;
    } else {
      ans[arr[i]] = -1;
      x1 -= p;
      y1 -= q;
    }
  }
  if (dist(x1, y1) <= MAXDIST) {
    return 1;
  }
  return 0;
}
signed main() {
  srand(time(0));
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    arr[i] = i;
    int p, q;
    cin >> pos[i].first >> pos[i].second;
  }
  while (!solve()) {
  }
  for (int i = 0; i < n; ++i) {
    cout << ans[i] << ' ';
  }
}