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 + 5;
const long long INF = 1e18;
const long long LIMIT = 1500000LL * 1500000LL;
vector<int> V;
int ans[MAXN];
struct node {
  long long x, y, pos;
  long long dis() { return x * x + y * y; }
};
node p[MAXN];
bool cmp(node A, node B) { return A.dis() > B.dis(); }
node operator+(const node &A, const node &B) {
  node ans = A;
  ans.x += B.x;
  ans.y += B.y;
  return ans;
}
node operator-(const node &A, const node &B) {
  node ans = A;
  ans.x -= B.x;
  ans.y -= B.y;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i].x >> p[i].y, p[i].pos = i;
  node cur{0, 0, 0};
  for (int i = n - 1; i >= 0; i--) {
    if (cmp(cur - p[i], cur + p[i]))
      cur = cur + p[i], ans[i] = 1;
    else
      cur = cur - p[i], ans[i] = -1;
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  return 0;
}