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 T = 1.5 * 1e6, N = 1e5 + 7;
long long dis(long long x, long long y) { return ((x) * (x)) + ((y) * (y)); }
long long x[N], y[N], X, Y;
int n, i, j, id[N], v[N];
int main() {
  for (scanf("%d", &n), i = 1; i <= n; ++i)
    scanf("%I64d%I64d", x + i, y + i), id[i] = i;
  for (;;) {
    for (random_shuffle(id + 1, id + n + 1), X = 0, Y = 0, i = 1; i <= n; ++i) {
      if (dis(X - x[id[i]], Y - y[id[i]]) < dis(X + x[id[i]], Y + y[id[i]])) {
        X -= x[id[i]];
        Y -= y[id[i]];
        v[id[i]] = -1;
      } else {
        X += x[id[i]];
        Y += y[id[i]];
        v[id[i]] = 1;
      }
    }
    if (dis(X, Y) > (long long)T * T) continue;
    for (i = 1; i <= n; ++i) printf("%d ", v[i]);
    return 0;
  }
}
