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;
void my_file_in_out() {
  freopen("inp.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
inline void OK(string t = "") { cerr << "OK " << t << '\n'; }
inline long double time() { return clock() / (long double)CLOCKS_PER_SEC; }
inline void stop(string t = "PROCESS FINISHED") {
  cerr << t << '\n';
  exit(0);
}
const long long M = 998244353;
const int MAXN = 2e5 + 10;
long long bp(long long c, long long st) {
  if (!st) return 1;
  if (st & 1) return (c * bp(c, st - 1)) % M;
  return bp((c * c) % M, st >> 1);
}
namespace VECTOR_MAGIC {
template <typename T>
istream &operator>>(istream &inp, vector<T> &a) {
  for (auto &elem : a) inp >> elem;
  return inp;
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> &a) {
  for (auto &elem : a) out << elem << ' ';
  out << endl;
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, vector<vector<T>> &a) {
  for (auto &elem : a) out << elem;
  return out;
}
template <typename T>
void operator++(vector<T> &a) {
  for (auto &el : a) el++;
}
template <typename T>
void operator++(vector<T> &a, T) {
  for (auto &el : a) el++;
}
template <typename T>
void operator--(vector<T> &a) {
  for (auto &el : a) el--;
}
template <typename T>
void operator--(vector<T> &a, T) {
  for (auto &el : a) el--;
}
template <typename T>
void operator+=(vector<T> &a, T X) {
  for (auto &el : a) el += X;
}
}  // namespace VECTOR_MAGIC
using namespace VECTOR_MAGIC;
mt19937 rnd(time(0));
const int INF = 1e9 + 7;
const long long LINF = 1e15 + 7;
struct vect {
  long long x, y, sum, mx;
  int id;
  vect() {}
  vect(long long x, long long y, int id = 0) {
    this->x = x;
    this->y = y;
    this->id = id;
    sum = abs(x) + abs(y);
    mx = max(abs(x), abs(y));
  }
};
vect operator-(vect &A, vect &B) { return {A.x - B.x, A.y - B.y}; }
vect operator+(vect &A, vect &B) { return {A.x + B.x, A.y + B.y}; }
long long dist(vect &A) { return A.x * A.x + A.y * A.y; }
signed main() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<vect> a(n);
  for (int i = 0; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    a[i] = vect(x, y, i);
  }
  sort(a.begin(), a.end(), [&](const vect &A, const vect &B) {
    if (A.sum != B.sum) return A.sum > B.sum;
    return A.mx > B.mx;
  });
  vector<int> ans(n);
  vect tmp(0, 0);
  for (int i = 0; i < n; i++) {
    vect t1 = tmp + a[i];
    vect t2 = tmp - a[i];
    if (dist(t1) < dist(t2)) {
      ans[a[i].id] = 1;
      tmp = t1;
    } else {
      ans[a[i].id] = -1;
      tmp = t2;
    }
  }
  cout << ans;
  return 0;
}