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 double EPS = 1e-9L;
const long double MAX_SQR_LEN = 2250000000000.0L;
bool eq(long double x, long double y) { return abs(x - y) < EPS; }
bool ls(long double x, long double y) { return x < y && !eq(x, y); }
bool lsEq(long double x, long double y) { return x < y || eq(x, y); }
struct Point {
  long double x, y;
  Point() : x(), y() {}
  Point(long double _x, long double _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 double& k) const { return Point(x * k, y * k); }
  Point operator/(const long double& k) const { return Point(x / k, y / k); }
  long double operator%(const Point& a) const { return x * a.x + y * a.y; }
  long double operator*(const Point& a) const { return x * a.y - y * a.x; }
  long double sqrLen() const { return x * x + y * y; }
  long double len() const { return sqrt(sqrLen()); }
  Point norm() const { return (*this) / len(); }
  Point ort() const { return Point(-y, x); }
  Point rot(const long double& cosA, const long double& sinA) const {
    return (*this) * cosA + ort() * sinA;
  }
  Point rot(const long double& ang) const { return rot(cos(ang), sin(ang)); }
  bool operator==(const Point& a) const { return eq(x, a.x) && eq(y, a.y); }
  bool operator<(const Point& a) const {
    if (!eq(x, a.x)) return x < a.x;
    return ls(y, a.y);
  }
};
int main() {
  srand(time(NULL));
  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 double distPlus = (pos + p[i].first).sqrLen();
      long double 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;
}