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;
struct Point {
  long long x, y;
  Point() {}
  Point(long long _x, long long _y) : x(_x), y(_y) {}
  void input() { scanf("%lld%lld", &x, &y); }
  Point operator+(const Point &b) const { return Point(x + b.x, y + b.y); }
  Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
  long long len() { return x * x + y * y; }
};
Point a[100010];
int res[100010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) a[i].input();
  if (n == 1) return 0 * printf("1\n");
  res[1] = 1;
  int j = 2;
  Point cur = a[1];
  for (int i = 3; i <= n; i++) {
    long long len1 = (cur + a[j]).len();
    long long len2 = (cur - a[j]).len();
    long long len3 = (cur + a[i]).len();
    long long len4 = (cur - a[i]).len();
    if (len1 <= len2 && len1 <= len3 && len1 <= len4) {
      cur = cur + a[j];
      res[j] = 1;
      j = i;
    } else if (len2 <= len1 && len2 <= len3 && len2 <= len4) {
      cur = cur - a[j];
      res[j] = -1;
      j = i;
    } else if (len3 <= len1 && len3 <= len2 && len3 <= len4) {
      cur = cur + a[i];
      res[i] = 1;
    } else {
      cur = cur - a[i];
      res[i] = -1;
    }
  }
  long long len1 = (cur + a[j]).len();
  long long len2 = (cur - a[j]).len();
  if (len1 <= len2)
    res[j] = 1;
  else
    res[j] = -1;
  for (int i = 1; i <= n; i++) printf("%d%c", res[i], i == n ? '\n' : ' ');
}