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 long long max_sqdis = 1000000000000LL;
pair<long long, long long> p[100001];
int sign[100001];
vector<int> sons[100001];
queue<int> que;
inline bool f(const int& a, const int& b, const int& c) {
  if ((p[a].first + p[b].first) * (p[a].first + p[b].first) +
          (p[a].second + p[b].second) * (p[a].second + p[b].second) <=
      max_sqdis) {
    p[a].first += p[b].first;
    p[a].second += p[b].second;
    sons[a].push_back(b);
    que.push(a);
    que.push(c);
    return true;
  }
  if ((p[a].first - p[b].first) * (p[a].first - p[b].first) +
          (p[a].second - p[b].second) * (p[a].second - p[b].second) <=
      max_sqdis) {
    p[a].first -= p[b].first;
    p[a].second -= p[b].second;
    sign[b] *= -1;
    sons[a].push_back(b);
    que.push(a);
    que.push(c);
    return true;
  }
  return false;
}
void dfs(const int& x) {
  for (vector<int>::iterator i = sons[x].begin(); i != sons[x].end(); ++i) {
    sign[*i] *= sign[x];
    dfs(*i);
  }
  return;
}
int main() {
  int i, n, a, b, c;
  scanf("%d", &n);
  for (i = 0; i < n; ++i) {
    scanf("%lld%lld", &p[i].first, &p[i].second);
    que.push(i);
    sons[i].clear();
    sign[i] = 1;
  }
  if (n == 1) {
    puts("1");
    return 0;
  }
  while (que.size() > 2) {
    a = que.front();
    que.pop();
    b = que.front();
    que.pop();
    c = que.front();
    que.pop();
    if (!f(a, b, c)) {
      if (!f(a, c, b)) {
        f(b, c, a);
      }
    }
  }
  a = que.front();
  que.pop();
  b = que.front();
  que.pop();
  if ((p[a].first + p[b].first) * (p[a].first + p[b].first) +
          (p[a].second + p[b].second) * (p[a].second + p[b].second) >
      max_sqdis << 1) {
    sign[b] *= -1;
  }
  sons[a].push_back(b);
  dfs(a);
  for (i = 0; i < n; ++i) {
    printf("%d ", sign[i]);
  }
  return 0;
}