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>
using namespace std;
const int LIM1 = 1000000;
const int LIM2 = 1500000;
int n, fa[100100], A, B;
bool rev[100100];
struct Vector {
  int x, y;
  Vector(int X = 0, int Y = 0) { x = X, y = Y; }
  friend Vector operator+(const Vector &u, const Vector &v) {
    return Vector(u.x + v.x, u.y + v.y);
  }
  friend Vector operator-(const Vector &u, const Vector &v) {
    return Vector(u.x - v.x, u.y - v.y);
  }
  friend bool operator<=(const Vector &u, const int &v) {
    return 1ll * u.x * u.x + 1ll * u.y * u.y <= 1ll * v * v;
  }
} v[100100];
int main() {
  scanf("%d", &n);
  if (n <= 1) {
    puts("1");
    return 0;
  }
  for (int i = 1; i <= n; i++) scanf("%d%d", &v[i].x, &v[i].y);
  A = 1, B = 2;
  for (int i = 3; i <= n; i++) {
    if (v[i] + v[A] <= LIM1) {
      v[i] = v[i] + v[A], fa[A] = i, A = B, B = i;
      continue;
    }
    if (v[i] - v[A] <= LIM1) {
      v[i] = v[i] - v[A], fa[A] = i, rev[A] ^= 1, A = B, B = i;
      continue;
    }
    if (v[i] + v[B] <= LIM1) {
      v[i] = v[i] + v[B], fa[B] = i, B = i;
      continue;
    }
    if (v[i] - v[B] <= LIM1) {
      v[i] = v[i] - v[B], fa[B] = i, rev[B] ^= 1, B = i;
      continue;
    }
    if (v[B] + v[A] <= LIM1) {
      v[B] = v[B] + v[A], fa[A] = B, A = B, B = i;
      continue;
    }
    if (v[B] - v[A] <= LIM1) {
      v[B] = v[B] - v[A], fa[A] = B, rev[A] ^= 1, A = B, B = i;
      continue;
    }
  }
  if (v[B] + v[A] <= LIM2)
    v[B] = v[B] + v[A], fa[A] = B;
  else
    v[B] = v[B] - v[A], fa[A] = B, rev[A] ^= 1;
  for (int i = n; i; i--)
    if (fa[i]) rev[i] ^= rev[fa[i]];
  for (int i = 1; i <= n; i++) printf("%d ", rev[i] ? -1 : 1);
  return 0;
}