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;
int n;
const long long LIM = 1500000;
const long long LIMC = 1000000;
const int N = 3e5;
long long a[N], b[N], d[N], ans[N];
bool cmp(const long long &i, const long long &j) {
  return (a[i] * a[i] + b[i] * b[i]) < (a[j] * a[j] + b[j] * b[j]);
}
long long sqr(long long x) { return x * x; }
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    d[i] = i;
  }
  sort(d + 1, d + n + 1, cmp);
  long long dx = 0, dy = 0;
  for (int i = 1; i <= n; i++) {
    long long mn = 1e18;
    int tp = 0;
    for (int t = -1; t <= 1; t += 2) {
      if (sqrt(sqr(dx + t * a[d[i]]) + sqr(dy + t * b[d[i]])) < mn) {
        mn = sqrt(sqr(dx + t * a[d[i]]) + sqr(dy + t * b[d[i]]));
        tp = t;
      }
    }
    dx += tp * a[d[i]];
    dy += tp * b[d[i]];
    ans[d[i]] = tp;
  }
  int it = 0;
  while (sqr(dx) + sqr(dy) > sqr(LIM)) {
    int l = rand() * rand() % n + 1;
    int r = l;
    if (r - l + 1 >= 1000) continue;
    if (l >= r) swap(l, r);
    long long sdx = dx, sdy = dy;
    for (int i = l; i <= r; i++) {
      dx -= ans[i] * a[i];
      dy -= ans[i] * b[i];
      ans[i] *= -1;
      dx += ans[i] * a[i];
      dy += ans[i] * b[i];
    }
    if (sqr(sdx) + sqr(sdy) < sqr(dx) + sqr(dy)) {
      for (int i = l; i <= r; i++) ans[i] *= -1;
      dx = sdx;
      dy = sdy;
    }
    it++;
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}