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 int N = 100005;
const long long M = 2250000000000;
int n;
pair<int, int> vec[N];
mt19937 myrand(time(NULL));
unsigned int rand(unsigned int l, unsigned int r) {
  return myrand() % (r - l + 1) + l;
}
int val[N];
bool check() {
  long long x = 0, y = 0;
  for (int i = 1; i <= n; i++) {
    x += val[i] * vec[i].first, y += val[i] * vec[i].second;
    if (x * x + y * y >= M)
      x -= val[i] * vec[i].first * 2, y -= val[i] * vec[i].second * 2,
          val[i] = -val[i];
  }
  return x * x + y * y <= M;
}
int main() {
  int st = clock();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    vec[i] = {x, y};
  }
  while (true) {
    if ((double)(clock() - st) / CLOCKS_PER_SEC > 1.9) {
      printf("-1");
      return 0;
    }
    for (int i = 1; i <= n; i++) val[i] = myrand() % 2 == 0 ? -1 : 1;
    if (check()) {
      for (int i = 1; i <= n; i++) printf("%d ", val[i]);
      return 0;
    }
  }
  return 0;
}