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;
int oo = (1e9) + 7;
long long oo2 = 1ll * oo * oo;
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, int> > v;
  vector<int> ord;
  for (int i = 0; i < n; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    v.push_back(make_pair(a, b));
    ord.push_back(i);
  }
  int cnt = 0;
  while (1) {
    cnt++;
    if (cnt == 3) assert(false);
    random_shuffle(ord.begin(), ord.end());
    vector<int> ans(n);
    ans[ord[0]] = 1;
    pair<int, int> pos = v[ord[0]];
    double d;
    for (int i = 1; i < n; ++i) {
      pair<int, int> op1;
      op1.first = v[ord[i]].first + pos.first;
      op1.second = v[ord[i]].second + pos.second;
      pair<int, int> op2;
      op2.first = -v[ord[i]].first + pos.first;
      op2.second = -v[ord[i]].second + pos.second;
      double d1 =
          sqrt(1ll * op1.first * op1.first + 1ll * op1.second * op1.second);
      double d2 =
          sqrt(1ll * op2.first * op2.first + 1ll * op2.second * op2.second);
      if (d1 < d2) {
        ans[ord[i]] = 1;
        pos = op1;
      } else {
        ans[ord[i]] = -1;
        pos = op2;
      }
    }
    if (sqrt(1ll * pos.first * pos.first + 1ll * pos.second * pos.second) >
        1.5e6)
      continue;
    for (int i = 0; i < n; ++i) {
      printf("%d ", ans[i]);
    }
    printf("\n");
    return 0;
  }
}