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;
using ll = long long;
using vi = vector<int>;
using vl = vector<long long>;
using pii = pair<int, int>;
using vpii = vector<pair<int, int>>;
template <class T>
inline bool ckmin(T& a, T b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
inline bool ckmax(T& a, T b) {
  return b > a ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const char nl = '\n';
const int mxN = 2e5 + 10;
const int MOD = 1e9 + 7;
const long long infLL = 1e18;
int n, x, y;
pair<pii, int> p[mxN];
int ans[mxN];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = (0); i < (n); i++) {
    cin >> x >> y;
    p[i] = {{x, y}, i};
  }
  const ll MX = 1LL * 1.5e6 * 1.5e6;
  while (1) {
    random_shuffle(p, p + n);
    ll x = 0, y = 0;
    for (int i = (n - 1); i >= (0); i--) {
      ll ux = x + p[i].first.first, uy = y + p[i].first.second;
      ll vx = x - p[i].first.first, vy = y - p[i].first.second;
      if (ux * ux + uy * uy < vx * vx + vy * vy) {
        ans[p[i].second] = 1, x = ux, y = uy;
      } else {
        ans[p[i].second] = -1, x = vx, y = vy;
      }
    }
    if (x * x + y * y <= MX) {
      for (int i = (0); i < (n); i++) cout << ans[i] << ' ';
      cout << nl;
      return 0;
    }
  }
  return 0;
}