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 N = 1e5 + 5, M = 1500000;
struct point {
  double len;
  int x, y, id;
  point() {}
  point(int _x, int _y, int _id) {
    x = _x;
    y = _y;
    id = _id;
    len = pow(1ll * x * x + 1ll * y * y, 0.5);
  }
  bool operator<(const point& b) const { return len > b.len; }
};
bool cmp1(point i, point j) { return i.x > j.x; }
bool cmp2(point i, point j) { return i.y > j.y; }
point a[N];
int ans[N];
void print(int n) {
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  exit(0);
}
double len(long long x, long long y) { return sqrt(x * x + y * y); }
int main() {
  long long nx = 0, ny = 0;
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x, y;
    cin >> x >> y;
    double tmp1 = len(nx + x, ny + y), tmp2 = len(nx - x, ny - y);
    if (tmp1 <= tmp2)
      nx += x, ny += y, ans[i] = 1;
    else
      nx -= x, ny -= y, ans[i] = -1;
    a[i] = point(x, y, i);
  }
  if (len(nx, ny) <= M) print(n);
  sort(a + 1, a + n + 1);
  nx = 0;
  ny = 0;
  for (int i = 1; i <= n; ++i) {
    int x = a[i].x, y = a[i].y;
    double tmp1 = len(nx + x, ny + y), tmp2 = len(nx - x, ny - y);
    if (tmp1 <= tmp2)
      nx += x, ny += y, ans[a[i].id] = 1;
    else
      nx -= x, ny -= y, ans[a[i].id] = -1;
  }
  if (len(nx, ny) <= M) print(n);
  sort(a + 1, a + n + 1, cmp1);
  nx = 0;
  ny = 0;
  for (int i = 1; i <= n; ++i) {
    int x = a[i].x, y = a[i].y;
    double tmp1 = len(nx + x, ny + y), tmp2 = len(nx - x, ny - y);
    if (tmp1 <= tmp2)
      nx += x, ny += y, ans[a[i].id] = 1;
    else
      nx -= x, ny -= y, ans[a[i].id] = -1;
  }
  if (len(nx, ny) <= M) print(n);
  sort(a + 1, a + n + 1, cmp2);
  nx = 0;
  ny = 0;
  for (int i = 1; i <= n; ++i) {
    int x = a[i].x, y = a[i].y;
    double tmp1 = len(nx + x, ny + y), tmp2 = len(nx - x, ny - y);
    if (tmp1 <= tmp2)
      nx += x, ny += y, ans[a[i].id] = 1;
    else
      nx -= x, ny -= y, ans[a[i].id] = -1;
  }
  print(n);
  return 0;
}