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;
const int N = 500500;
const int mod = 1e9 + 7;
const long long inf = 1e18;
int x[N], y[N];
int n;
int ans[N];
inline long long calc(long long x, long long y) { return x * x + y * y; }
int id[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  srand(time(NULL));
  for (long long i = 1; i <= n; ++i) id[i] = i;
  for (long long i = 1; i <= n; ++i) cin >> x[i] >> y[i], id[i] = i;
  do {
    int X = 0, Y = 0;
    for (long long i = 1; i <= n; ++i) {
      int nx1 = abs(X + x[i]);
      int ny1 = abs(Y + y[i]);
      int nx2 = abs(X - x[i]);
      int ny2 = abs(Y - y[i]);
      if (max(nx1, ny1) <= max(nx2, ny2)) {
        X += x[i];
        Y += y[i];
        ans[id[i]] = 1;
      } else {
        X -= x[i];
        Y -= y[i];
        ans[id[i]] = -1;
      }
    }
    if (calc(X, Y) <= 1500000ll * 1500000) {
      for (long long i = 1; i <= n; ++i) cout << ans[i] << ' ';
      return 0;
    }
    for (long long i = 1; i <= n; ++i) {
      int j = rand() % n + 1;
      swap(x[j], x[i]);
      swap(y[i], y[j]);
      swap(id[i], id[j]);
    }
  } while (true);
  return 0;
}
