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 double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
int n;
long long x[100015], y[100015];
struct gg {
  long long x, y;
  int index;
  gg() {}
  gg(long long x, long long y, int index) : x(x), y(y), index(index) {}
} g[100015];
long long pp(long long x) { return x * x; }
int ans[100015];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld%lld", &x[i], &y[i]);
    g[i] = gg(x[i], y[i], i);
  }
  while (1) {
    random_shuffle(g + 1, g + n + 1);
    long long tx = 0, ty = 0;
    for (int i = 1; i <= n; ++i) {
      if (pp(tx + g[i].x) + pp(ty + g[i].y) <
          pp(tx - g[i].x) + pp(ty - g[i].y)) {
        ans[g[i].index] = 1;
        tx += g[i].x, ty += g[i].y;
      } else {
        ans[g[i].index] = -1;
        tx -= g[i].x, ty -= g[i].y;
      }
    }
    if (tx * tx + ty * ty <= 2.25e12) break;
  }
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  return 0;
}
