#include <bits/stdc++.h>
std::mt19937 rnd(time(0));
const int N = 100005;
const long long LIM = 1500000ll * 1500000ll;
int n;
struct Node {
  int x, y;
  Node() { x = 0, y = 0; }
  Node(int _x, int _y) : x(_x), y(_y) {}
  Node operator+(const Node &rhs) const { return Node(x + rhs.x, y + rhs.y); }
  Node operator*(const int t) const { return Node(x * t, y * t); }
  long long len2() { return 1ll * x * x + 1ll * y * y; }
};
Node a[N];
int p[N];
int ans[N];
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  std::cin >> n;
  for (int i = 1; i <= n; ++i) {
    std::cin >> a[i].x >> a[i].y;
  }
  while (1) {
    for (int i = 1; i <= n; ++i) {
      p[i] = i;
    }
    std::shuffle(p + 1, p + 1 + n, rnd);
    Node now(0, 0);
    for (int i = 1; i <= n; ++i) {
      Node nx1 = now + a[p[i]], nx2 = now + a[p[i]] * (-1);
      if ((nx1.len2() <= LIM) == (nx2.len2() <= LIM)) {
        if (rnd() & 1) {
          now = nx1;
          ans[p[i]] = 1;
        } else {
          now = nx2;
          ans[p[i]] = -1;
        }
      } else if (nx1.len2() <= LIM) {
        now = nx1;
        ans[p[i]] = 1;
      } else {
        now = nx2;
        ans[p[i]] = -1;
      }
    }
    if (now.len2() <= LIM) {
      for (int i = 1; i <= n; ++i) {
        std::cout << ans[i] << " \n"[i == n];
      }
      return 0;
    }
  }
}

### Prompt

Generate 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>
std::mt19937 rnd(time(0));
const int N = 100005;
const long long LIM = 1500000ll * 1500000ll;
int n;
struct Node {
  int x, y;
  Node() { x = 0, y = 0; }
  Node(int _x, int _y) : x(_x), y(_y) {}
  Node operator+(const Node &rhs) const { return Node(x + rhs.x, y + rhs.y); }
  Node operator*(const int t) const { return Node(x * t, y * t); }
  long long len2() { return 1ll * x * x + 1ll * y * y; }
};
Node a[N];
int p[N];
int ans[N];
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  std::cin >> n;
  for (int i = 1; i <= n; ++i) {
    std::cin >> a[i].x >> a[i].y;
  }
  while (1) {
    for (int i = 1; i <= n; ++i) {
      p[i] = i;
    }
    std::shuffle(p + 1, p + 1 + n, rnd);
    Node now(0, 0);
    for (int i = 1; i <= n; ++i) {
      Node nx1 = now + a[p[i]], nx2 = now + a[p[i]] * (-1);
      if ((nx1.len2() <= LIM) == (nx2.len2() <= LIM)) {
        if (rnd() & 1) {
          now = nx1;
          ans[p[i]] = 1;
        } else {
          now = nx2;
          ans[p[i]] = -1;
        }
      } else if (nx1.len2() <= LIM) {
        now = nx1;
        ans[p[i]] = 1;
      } else {
        now = nx2;
        ans[p[i]] = -1;
      }
    }
    if (now.len2() <= LIM) {
      for (int i = 1; i <= n; ++i) {
        std::cout << ans[i] << " \n"[i == n];
      }
      return 0;
    }
  }
}

```