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 vec {
  int id;
  int x, y;
  long long lng;
  inline vec operator+(const vec& foo) {
    vec res;
    res.x = x + foo.x;
    res.y = y + foo.y;
    res.id = 0;
    return res;
  }
  inline vec operator-(const vec& foo) {
    vec res;
    res.x = x - foo.x;
    res.y = y - foo.y;
    res.id = 0;
    return res;
  }
  void calc_lng() { lng = (x * 1LL * x + y * 1LL * y); }
};
const int MX = 100000 + 5;
const long long MAX = 1500000 * 1LL * 1500000;
int color[MX];
int X[MX], Y[MX];
bool solve(int n, const vector<vec>& v) {
  vec all;
  all.x = 0;
  all.y = 0;
  int mult = 1;
  for (auto x : v) {
    vec One = all + x;
    vec Two = all - x;
    vec Three = x - all;
    One.calc_lng();
    Two.calc_lng();
    Three.calc_lng();
    if (One.lng <= Two.lng and One.lng <= Two.lng) {
      color[x.id] = +1 * mult;
      all = One;
    } else if (Two.lng <= One.lng and Two.lng <= Three.lng) {
      color[x.id] = -1 * mult;
      all = Two;
    } else if (Three.lng <= One.lng and Three.lng <= Two.lng) {
      mult *= -1;
      color[x.id] = mult;
      all = Three;
    }
  }
  all.calc_lng();
  if (all.lng > MAX) return false;
  for (int i = 0; i < n; i++) {
    printf("%d ", color[i] * mult);
  }
  printf("\n");
  return true;
}
int main() {
  int n;
  ignore = scanf("%d", &n);
  vector<vec> v;
  for (int i = 0; i < n; i++) {
    vec tmp;
    tmp.id = i;
    ignore = scanf("%d %d", &tmp.x, &tmp.y);
    v.push_back(tmp);
    X[i] = tmp.x;
    Y[i] = tmp.y;
  }
  while (true) {
    if (solve(n, v)) break;
    random_shuffle(v.begin(), v.end());
  }
}