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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long MAX_SQR_LEN = 2250000000000L;
bool eq(long long x, long long y) { return x == y; }
bool lsEq(long long x, long long y) { return x <= y; }
struct Point {
  long long x, y;
  Point() : x(), y() {}
  Point(long long _x, long long _y) : x(_x), y(_y) {}
  Point operator+(const Point& a) const { return Point(x + a.x, y + a.y); }
  Point operator-(const Point& a) const { return Point(x - a.x, y - a.y); }
  Point operator*(const long long& k) const { return Point(x * k, y * k); }
  long long sqrLen() const { return x * x + y * y; }
};
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<pair<Point, int>> p(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i].first.x >> p[i].first.y;
    p[i].second = i;
  }
  mt19937 randGen(time(0));
  vector<int> c;
  while (true) {
    shuffle(p.begin(), p.end(), randGen);
    c.assign(n, 0);
    Point pos(0, 0);
    for (int i = 0; i < n; i++) {
      long long distPlus = (pos + p[i].first).sqrLen();
      long long distMinus = (pos - p[i].first).sqrLen();
      if (lsEq(distPlus, distMinus)) {
        c[p[i].second] = 1;
        pos = pos + p[i].first;
      } else {
        c[p[i].second] = -1;
        pos = pos - p[i].first;
      }
    }
    if (lsEq(pos.sqrLen(), MAX_SQR_LEN)) break;
  }
  for (int i = 0; i < n; i++) cout << c[i] << " ";
  cout << endl;
}