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;
long long mod = 1e9 + 7;
const long long mx_error = 15 * 1e5;
long long px = 0, py = 0;
bool try_it(long long dx, long long dy) {
  long long n_x = px + dx;
  long long n_y = py + dy;
  long long n_x_neg = px - dx;
  long long n_y_neg = py - dy;
  long long er = n_x * n_x + n_y * n_y;
  long long er_neg = n_x_neg * n_x_neg + n_y_neg * n_y_neg;
  if (er > er_neg) return false;
  return true;
}
struct delta {
  int idx;
  long long dx, dy;
};
bool cmp(delta a, delta b) { return a.dx > b.dx; }
int ans[100009];
int main() {
  int n;
  scanf("%d", &n);
  vector<delta> a(n);
  for (int i = 0; i < n; i++) {
    int dx, dy;
    scanf("%d %d", &dx, &dy);
    a[i] = {i, dx, dy};
  }
  bool done = false;
  while (!done) {
    px = 0, py = 0;
    random_shuffle(a.begin(), a.end());
    done = true;
    for (int i = 0; i < n; i++) {
      long long dx, dy;
      dx = a[i].dx;
      dy = a[i].dy;
      int res = 1;
      if (!try_it(dx, dy)) res = -1;
      px += 1LL * res * dx;
      py += 1LL * res * dy;
      ans[a[i].idx] = res;
    }
    if (px > mx_error || py > mx_error ||
        px * px + py * py > mx_error * mx_error)
      done = false;
  }
  for (int i = 0; i < n; i++) printf("%d%c", ans[i], " \n"[i == n - 1]);
  cerr << "final:" << px << ' ' << py << '\n';
  cerr << "final error = " << px * px + py * py << '\n';
}