#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
struct vec {
  int x, y, i;
  vec(int _x = 0, int _y = 0) : x(_x), y(_y), i(0) {}
  vec operator+(const vec& o) const { return vec(x + o.x, y + o.y); }
  vec operator+=(const vec& o) {
    x += o.x;
    y += o.y;
    return *this;
  }
  vec operator-() const { return vec(-x, -y); }
  vec operator-(const vec& o) const { return *this + (-o); }
  vec operator-=(const vec& o) {
    x -= o.x;
    y -= o.y;
    return *this;
  }
};
const long long TARGET = 1.5e6 * 1.5e6;
long long abs(const vec& v) {
  return (long long)v.x * v.x + (long long)v.y * v.y;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<vec> vecs(n);
  for (int i = 0; i < n; i++) {
    cin >> vecs[i].x >> vecs[i].y;
    vecs[i].i = i;
  }
  vector<int> ans(n);
  while (1) {
    shuffle(vecs.begin(), vecs.end(),
            mt19937(chrono::system_clock::now().time_since_epoch().count()));
    vec result;
    for (const auto& v : vecs) {
      if (abs(result + v) < abs(result - v)) {
        result += v;
        ans[v.i] = 1;
      } else {
        result -= v;
        ans[v.i] = -1;
      }
    }
    if (abs(result) <= TARGET) {
      for (int i : ans) {
        cout << i << " ";
      }
      cout << '\n';
      return 0;
    }
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following 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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
struct vec {
  int x, y, i;
  vec(int _x = 0, int _y = 0) : x(_x), y(_y), i(0) {}
  vec operator+(const vec& o) const { return vec(x + o.x, y + o.y); }
  vec operator+=(const vec& o) {
    x += o.x;
    y += o.y;
    return *this;
  }
  vec operator-() const { return vec(-x, -y); }
  vec operator-(const vec& o) const { return *this + (-o); }
  vec operator-=(const vec& o) {
    x -= o.x;
    y -= o.y;
    return *this;
  }
};
const long long TARGET = 1.5e6 * 1.5e6;
long long abs(const vec& v) {
  return (long long)v.x * v.x + (long long)v.y * v.y;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<vec> vecs(n);
  for (int i = 0; i < n; i++) {
    cin >> vecs[i].x >> vecs[i].y;
    vecs[i].i = i;
  }
  vector<int> ans(n);
  while (1) {
    shuffle(vecs.begin(), vecs.end(),
            mt19937(chrono::system_clock::now().time_since_epoch().count()));
    vec result;
    for (const auto& v : vecs) {
      if (abs(result + v) < abs(result - v)) {
        result += v;
        ans[v.i] = 1;
      } else {
        result -= v;
        ans[v.i] = -1;
      }
    }
    if (abs(result) <= TARGET) {
      for (int i : ans) {
        cout << i << " ";
      }
      cout << '\n';
      return 0;
    }
  }
  return 0;
}

```