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>
using namespace std;
const int N = 1e5 + 5, lim = 1.5e6;
mt19937 mt(time(0) + (size_t) new int);
struct point {
  long long x, y;
  inline point operator+(const point& rhs) const {
    return (point){x + rhs.x, y + rhs.y};
  }
  inline point operator-(const point& rhs) const {
    return (point){x - rhs.x, y - rhs.y};
  }
  inline double length() { return sqrt(x * x + y * y); }
} a[N], p;
int n, fh[N];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].x >> a[i].y;
    fh[i] = 2 * int(mt() % 2) - 1;
    a[i].x *= fh[i], a[i].y *= fh[i];
    p = p + a[i];
  }
  while (p.length() > lim) {
    for (int T = 1; T <= 100000; ++T) {
      int t = mt() % n + 1;
      point q = p - a[t] - a[t];
      if (q.length() < p.length()) {
        a[t].x *= -1, a[t].y *= -1;
        fh[t] *= -1;
        p = q;
      }
      if (p.length() <= lim) break;
    }
    if (p.length() <= lim) break;
    p = (point){0, 0};
    for (int i = 1; i <= n; ++i) {
      a[i].x *= fh[i], a[i].y *= fh[i];
      fh[i] = 2 * int(mt() % 2) - 1;
      a[i].x *= fh[i], a[i].y *= fh[i];
      p = p + a[i];
    }
  }
  for (int i = 1; i <= n; ++i) cout << fh[i] << ' ';
  return 0;
}
