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;
const long long maxx = 2.25e12;
struct v {
  long long x, y;
  int id;
} vs[100005];
int n;
int ans[100005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld%lld", &vs[i].x, &vs[i].y);
    vs[i].id = i;
  }
  long long sumx = 0;
  long long sumy = 0;
  while (1) {
    sumx = 0;
    sumy = 0;
    random_shuffle(vs + 1, vs + 1 + n);
    for (int i = 1; i <= n; i++) {
      long long tmp1 = (sumx + vs[i].x) * (sumx + vs[i].x) +
                       (sumy + vs[i].y) * (sumy + vs[i].y);
      long long tmp2 = (sumx - vs[i].x) * (sumx - vs[i].x) +
                       (sumy - vs[i].y) * (sumy - vs[i].y);
      if (tmp1 > tmp2) {
        sumx -= vs[i].x;
        sumy -= vs[i].y;
        ans[vs[i].id] = -1;
      } else {
        sumx += vs[i].x;
        sumy += vs[i].y;
        ans[vs[i].id] = 1;
      }
    }
    if (sumx * sumx + sumy * sumy <= maxx) {
      break;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}