#include <bits/stdc++.h>
struct PT {
  long long x, y;
  int idx;
  bool get;
  PT(long long x = 0, long long y = 0) : x(x), y(y) { get = false; }
  PT operator+(const PT &p) const { return PT(x + p.x, y + p.y); }
  PT operator-(const PT &p) const { return PT(x - p.x, y - p.y); }
  PT operator*(long long c) const { return PT(x * c, y * c); }
  PT operator/(double c) const { return PT(x / c, y / c); }
  long long operator*(const PT &p) const { return x * p.x + y * p.y; }
  long long operator%(const PT &p) const { return x * p.y - y * p.x; }
  long long operator!() const { return *this * *this; }
};
PT eval(std::vector<PT> &pts) {
  PT p;
  for (auto pt : pts) {
    if (pt.get) {
      p = p + pt;
    } else {
      p = p - pt;
    }
  }
  return p;
}
int main() {
  srand(time(0));
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int n;
  std::cin >> n;
  std::vector<PT> pts(n);
  for (int i = 0; i < n; i++) {
    std::cin >> pts[i].x >> pts[i].y;
    pts[i].idx = i;
    if (i == 1) pts[i].get = false;
  }
  const long long INF = 1.5e6;
  while (!(eval(pts)) > INF * INF) {
    PT p;
    for (int i = 0; i < n; i++) {
      auto &pt = pts[i];
      if (!(p + pt) <= !(p - pt)) {
        p = p + pt;
        pt.get = true;
      } else {
        p = p - pt;
        pt.get = false;
      }
    }
    std::random_shuffle(pts.begin(), pts.end());
  }
  std::vector<bool> ans(n);
  for (auto p : pts) {
    ans[p.idx] = p.get;
  }
  for (int i = 0; i < n; i++) {
    std::cout << (ans[i] ? 1 : -1) << (i + 1 == n ? '\n' : ' ');
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 

### Response

```cpp
#include <bits/stdc++.h>
struct PT {
  long long x, y;
  int idx;
  bool get;
  PT(long long x = 0, long long y = 0) : x(x), y(y) { get = false; }
  PT operator+(const PT &p) const { return PT(x + p.x, y + p.y); }
  PT operator-(const PT &p) const { return PT(x - p.x, y - p.y); }
  PT operator*(long long c) const { return PT(x * c, y * c); }
  PT operator/(double c) const { return PT(x / c, y / c); }
  long long operator*(const PT &p) const { return x * p.x + y * p.y; }
  long long operator%(const PT &p) const { return x * p.y - y * p.x; }
  long long operator!() const { return *this * *this; }
};
PT eval(std::vector<PT> &pts) {
  PT p;
  for (auto pt : pts) {
    if (pt.get) {
      p = p + pt;
    } else {
      p = p - pt;
    }
  }
  return p;
}
int main() {
  srand(time(0));
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int n;
  std::cin >> n;
  std::vector<PT> pts(n);
  for (int i = 0; i < n; i++) {
    std::cin >> pts[i].x >> pts[i].y;
    pts[i].idx = i;
    if (i == 1) pts[i].get = false;
  }
  const long long INF = 1.5e6;
  while (!(eval(pts)) > INF * INF) {
    PT p;
    for (int i = 0; i < n; i++) {
      auto &pt = pts[i];
      if (!(p + pt) <= !(p - pt)) {
        p = p + pt;
        pt.get = true;
      } else {
        p = p - pt;
        pt.get = false;
      }
    }
    std::random_shuffle(pts.begin(), pts.end());
  }
  std::vector<bool> ans(n);
  for (auto p : pts) {
    ans[p.idx] = p.get;
  }
  for (int i = 0; i < n; i++) {
    std::cout << (ans[i] ? 1 : -1) << (i + 1 == n ? '\n' : ' ');
  }
}

```