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 
#include <bits/stdc++.h>
using namespace std;
template <class T>
using min_queue = priority_queue<T, vector<T>, greater<T>>;
template <typename Args>
void kill(Args args) {
  cout << args << "\n";
  exit(0);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const double PI = acos(-1);
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const int N = 201010;
int idx[N];
long long x[N], y[N], ans[N];
int n;
void solve(int tc) {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    idx[i] = i;
    cin >> x[i];
    cin >> y[i];
    if (x[i] < 0) {
      x[i] *= -1;
      y[i] *= -1;
      ans[i] ^= 1;
    }
  }
  sort(idx + 1, idx + n + 1,
       [](int i, int j) { return y[i] * x[j] < y[j] * x[i]; });
  long long X = x[idx[1]], Y = y[idx[1]];
  for (int i = 2; i <= n; i++) {
    int j = idx[i];
    long double mag1 = sqrtl((X - x[j]) * (X - x[j]) + (Y - y[j]) * (Y - y[j]));
    long double mag2 = sqrtl((X + x[j]) * (X + x[j]) + (Y + y[j]) * (Y + y[j]));
    if (mag1 < mag2) {
      ans[j] ^= 1;
      X = X - x[j];
      Y = Y - y[j];
    } else {
      X = X + x[j];
      Y = Y + y[j];
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << (ans[i] ? -1 : 1) << " \n"[i == n];
  }
}
void precompute() {}
int main() {
  cout << fixed << setprecision(12);
  int testcase = 1;
  clock_t zzzx = clock();
  precompute();
  for (int i = 1; i <= testcase; i++) {
    solve(i);
  }
  double elapsed_time = (double)(clock() - zzzx) / CLOCKS_PER_SEC;
  cerr << "elapsed_time"
       << " = " << (elapsed_time) << "\n";
  return 0;
}
