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 vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;
using vpii = vector<pair<int, int>>;
using vpll = vector<pair<ll, ll>>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const ll LIM = 1500000LL * 1500000LL;
struct xy_move {
  int x, y;
  int index;
};
int n;
vector<xy_move> moves;
vb flip;
ll sqdist(ll x, ll y) { return x * x + y * y; }
bool check() {
  ll x = 0, y = 0;
  for (auto m : moves) {
    ll x1 = x + m.x, y1 = y + m.y, x2 = x - m.x, y2 = y - m.y;
    if (sqdist(x1, y1) < sqdist(x2, y2)) {
      x = x1;
      y = y1;
      flip[m.index] = false;
    } else {
      x = x2;
      y = y2;
      flip[m.index] = true;
    }
  }
  return abs(x) <= 15e5 && abs(y) <= 15e5 && sqdist(x, y) <= LIM;
}
void solve() {
  int n;
  cin >> n;
  moves.resize(n);
  flip.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> moves[i].x >> moves[i].y;
    moves[i].index = i;
  }
  do {
    random_shuffle(moves.begin(), moves.end());
  } while (!check());
  for (int i = 0; i < n; i++) {
    if (flip[i])
      cout << -1 << " ";
    else
      cout << 1 << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t = 1;
  while (t--) {
    solve();
  }
}