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 maxn = 1e5 + 10;
struct VECTOR {
  long long x, y;
  int id;
} v[maxn];
int ans[maxn];
long long get(long long x, long long y) { return x * x + y * y; }
int main() {
  int n;
  srand((unsigned)time(NULL));
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      scanf("%lld%lld", &v[i].x, &v[i].y);
      v[i].id = i;
    }
    while (1) {
      long long x = v[0].x, y = v[0].y;
      ans[v[0].id] = 1;
      for (int i = 1; i < n; i++) {
        long long tx = x + v[i].x;
        long long ty = y + v[i].y;
        long long dis = get(tx, ty);
        tx = x - v[i].x;
        ty = y - v[i].y;
        if (get(tx, ty) <= dis) {
          x = x - v[i].x;
          y = y - v[i].y;
          ans[v[i].id] = -1;
          continue;
        }
        x = x + v[i].x;
        y = y + v[i].y;
        ans[v[i].id] = 1;
      }
      if (get(x, y) <= 2250000000000ll) {
        break;
      } else
        random_shuffle(v, v + n);
    }
    for (int i = 0; i < n; i++) {
      printf("%d", ans[i]);
      if (i != n - 1) putchar(' ');
    }
    putchar('\n');
  }
  return 0;
}
