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 mod = (long long)1e9 + 7;
const long long inf = (long long)1e16;
const long double eps = 1e-12;
const long long N = (int)1e5 + 5;
const long long LOGN = 19;
const long double PI = 3.14159265358979323846;
inline long long mul(long long a, long long b, long long m = mod) {
  return (long long)(a * b) % m;
}
inline long long add(long long a, long long b, long long m = mod) {
  a += b;
  if (a >= m) a -= m;
  if (a < 0) a += m;
  return a;
}
inline long long power(long long a, long long b, long long m = mod) {
  if (b == 0) return 1;
  if (b == 1) return (a % m);
  long long x = power(a, b / 2, m);
  x = mul(x, x, m);
  if (b % 2) x = mul(x, a, m);
  return x;
}
long long dist(long long x, long long y) { return x * x + y * y; }
long long chk = (long long)1.5e6 * (long long)1.5e6 * 1ll;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (0) {
    freopen("C:\\Users\\Dishant\\Desktop\\Collection-DEV c++\\input.txt", "r",
            stdin);
    freopen("C:\\Users\\Dishant\\Desktop\\Collection-DEV c++\\output.txt", "w",
            stdout);
  }
  int n;
  cin >> n;
  int ans[n];
  pair<int, pair<int, int> > p[n];
  for (int i = 0; i < n; i++) {
    p[i].first = i;
    cin >> p[i].second.first >> p[i].second.second;
  }
  srand(time(NULL));
  while (1) {
    long long curx = 0, cury = 0;
    random_shuffle(p, p + n);
    for (int i = 0; i < n; i++) {
      if (dist(curx + p[i].second.first, cury + p[i].second.second) <
          dist(curx - p[i].second.first, cury - p[i].second.second)) {
        ans[p[i].first] = 1;
        curx += p[i].second.first;
        cury += p[i].second.second;
      } else {
        ans[p[i].first] = -1;
        curx -= p[i].second.first;
        cury -= p[i].second.second;
      }
    }
    long long dis = dist(curx, cury);
    if (dis <= chk) break;
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  return 0;
}