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;
mt19937 ran(time(nullptr));
long long vx[100010], vy[100010];
int ord[100010], ans[100010];
inline int rnd(int p) { return ran() % p; }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld%lld", vx + i, vy + i);
    ord[i] = i;
  }
  bool found = false;
  while (!found) {
    random_shuffle(ord + 1, ord + 1 + n, rnd);
    long long cur_x = vx[ord[1]], cur_y = vy[ord[1]];
    ans[ord[1]] = 1;
    for (int i = 2; i <= n; i++) {
      long long nx1 = cur_x + vx[ord[i]], ny1 = cur_y + vy[ord[i]];
      long long nx2 = cur_x - vx[ord[i]], ny2 = cur_y - vy[ord[i]];
      if (nx1 * nx1 + ny1 * ny1 < nx2 * nx2 + ny2 * ny2) {
        cur_x = nx1, cur_y = ny1;
        ans[ord[i]] = 1;
      } else {
        cur_x = nx2, cur_y = ny2;
        ans[ord[i]] = -1;
      }
    }
    if (cur_x * cur_x + cur_y * cur_y <= 2250000000000ll) found = true;
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}