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;
struct vv {
  int x;
  int y;
  int id;
};
vv v[100005];
int ans[100005];
int main() {
  srand(time(0));
  int n;
  scanf("%d", &n);
  int nowx = 0, nowy = 0;
  int i;
  for (i = 0; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[i].id = i;
    v[i].x = x;
    v[i].y = y;
    long long t1 =
        (long long)(nowx - x) * (nowx - x) + (long long)(nowy - y) * (nowy - y);
    long long t2 =
        (long long)(nowx + x) * (nowx + x) + (long long)(nowy + y) * (nowy + y);
    if (t1 < t2) {
      ans[i] = -1;
      nowx -= x;
      nowy -= y;
    } else {
      ans[i] = 1;
      nowx += x;
      nowy += y;
    }
  }
  if ((long long)nowx * nowx + (long long)nowy * nowy >
      (long long)1500000 * 1500000) {
    for (;;) {
      random_shuffle(v, v + n);
      int nowx = 0, nowy = 0;
      for (i = 0; i < n; i++) {
        int x = v[i].x;
        int y = v[i].y;
        long long t1 = (long long)(nowx - x) * (nowx - x) +
                       (long long)(nowy - y) * (nowy - y);
        long long t2 = (long long)(nowx + x) * (nowx + x) +
                       (long long)(nowy + y) * (nowy + y);
        if (t1 < t2) {
          ans[v[i].id] = -1;
          nowx -= x;
          nowy -= y;
        } else {
          ans[v[i].id] = 1;
          nowx += x;
          nowy += y;
        }
      }
      if ((long long)nowx * nowx + (long long)nowy * nowy <=
          (long long)1500000 * 1500000) {
        for (i = 0; i < n; i++) {
          printf("%d ", ans[i]);
        }
        printf("\n");
        return 0;
      }
    }
  }
  for (i = 0; i < n; i++) {
    printf("%d ", ans[i]);
  }
  printf("\n");
  return 0;
}