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 maxn = 1e5 + 100;
struct node {
  int id;
  long long x, y;
} data[maxn];
int ans[maxn];
int main() {
  srand(time(NULL));
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> data[i].x >> data[i].y;
    data[i].id = i;
  }
  while (true) {
    long long tmp = 0, t1, t2;
    long long prex = data[1].x, prey = data[1].y;
    ans[data[1].id] = 1;
    for (int i = 2; i <= n; i++) {
      long long tx1 = prex - data[i].x;
      long long ty1 = prey - data[i].y;
      t1 = tx1 * tx1 + ty1 * ty1;
      long long tx2 = data[i].x + prex;
      long long ty2 = data[i].y + prey;
      t2 = tx2 * tx2 + ty2 * ty2;
      if (t1 < t2) {
        ans[data[i].id] = -1;
        prex = tx1;
        prey = ty1;
      } else {
        ans[data[i].id] = 1;
        prex = tx2;
        prey = ty2;
      }
    }
    if (min(t1, t2) <= 2250000000000LL) {
      break;
    } else {
      random_shuffle(data + 1, data + n + 1);
    }
  }
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], i == n ? '\n' : ' ');
}