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;
const int maxn = 1e5 + 100;
int ans[maxn];
struct Point {
  long long x, y, id;
  Point(long long xx = 0, long long yy = 0, long long iid = 0) {
    x = xx;
    y = yy;
    id = iid;
  }
  friend Point operator+(const Point &a, const Point &b) {
    return Point(a.x + b.x, a.y + b.y);
  }
  friend Point operator-(const Point &a, const Point &b) {
    return Point(a.x - b.x, a.y - b.y);
  }
} a[maxn];
long long get_mod(Point a) { return a.x * a.x + a.y * a.y; }
int main() {
  int n;
  srand(time(NULL));
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%I64d%I64d", &a[i].x, &a[i].y);
    a[i].id = i;
  }
  random_shuffle(a, a + n);
  Point now = Point(0, 0);
  for (int i = 0; i < n; i++) {
    if (get_mod(now + a[i]) < get_mod(now - a[i])) {
      ans[a[i].id] = 1;
      now = now + a[i];
    } else {
      ans[a[i].id] = -1;
      now = now - a[i];
    }
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", ans[i]);
  }
  printf("\n");
}