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 p = 2250000000000ll;
const long long N = 100005;
long long n;
int32_t dir[N];
int32_t a[2][N];
long long flag = 0;
void dp(long long in, long long x, long long y) {
  if (in == n + 1 && x * x + y * y <= p) flag = 1;
  if (in == n + 1 && !flag) return;
  if (flag == 1) {
    for (long long i = 1; i <= n; i++) cout << dir[i] << ' ';
    exit(0);
  }
  long long ax = x - a[0][in];
  long long ay = y - a[1][in];
  long long bx = x + a[0][in];
  long long by = y + a[1][in];
  if (ax * ax + ay * ay < bx * bx + by * by) {
    dir[in] = -1;
    dp(in + 1, x - a[0][in], y - a[1][in]);
    dir[in] = 1;
    dp(in + 1, x + a[0][in], y + a[1][in]);
  } else {
    dir[in] = 1;
    dp(in + 1, x + a[0][in], y + a[1][in]);
    dir[in] = -1;
    dp(in + 1, x - a[0][in], y - a[1][in]);
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[0][i] >> a[1][i];
  dp(1, 0, 0);
  for (long long i = 1; i <= n; i++) cout << dir[i] << ' ';
  return 0;
}