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;
long long int i, n;
long double x[200000], y[200000];
void nhap() {
  cin >> n;
  for (i = 1; i <= n; i++) cin >> x[i] >> y[i];
}
bool checked(long double px, long double py, long long int i) {
  int dem = 1, tmp, luu = 0;
  long double px1, py1, x1, y1;
  px1 = px;
  py1 = py;
  x1 = x[i];
  y1 = y[i];
  if ((px == 0) || (x[i] == 0)) {
    if ((py < 0) && (y[i] < 0)) return true;
    if ((py > 0) && (y[i] > 0)) return true;
  }
  if ((py == 0) || (y[i] == 0)) {
    if ((px < 0) && (x[i] < 0)) return true;
    if ((px > 0) && (x[i] > 0)) return true;
  }
  if ((px > 0) && (x[i] > 0) && (py > 0) && (y[i] > 0)) return true;
  if ((px < 0) && (x[i] < 0) && (py < 0) && (y[i] < 0)) return true;
  if ((px != 0) && (x[i] != 0) && (py != 0) && (y[i] != 0)) {
    if (y[i] < 0) {
      dem = dem + 1;
      y1 = abs(y[i]);
    }
    if (py < 0) {
      dem = dem + 1;
      py1 = abs(py);
    }
    if (px < 0) {
      dem = dem - 1;
      px1 = abs(px);
      luu = luu + 1;
    }
    if (x[i] < 0) {
      dem = dem - 1;
      x1 = abs(x[i]);
      luu = luu + 1;
    }
    dem = abs(dem);
    if (dem % 2 == 0)
      tmp = 1;
    else
      tmp = -1;
    if (luu % 2 == 0)
      if ((max(px1, x1) / max(py1, y1)) * (min(px1, x1) / min(py1, y1)) > tmp)
        return true;
    if (luu % 2 != 0)
      if ((max(px1, x1) / max(py1, y1)) * (min(px1, x1) / min(py1, y1)) < tmp)
        return true;
  }
  return false;
}
void check() {
  long double px, py;
  short int nho[200000];
  px = 0;
  py = 0;
  for (i = n; i >= 1; i--) {
    if (checked(px, py, i) == true) {
      nho[i] = -1;
      px = px - x[i];
      py = py - y[i];
    } else {
      nho[i] = 1;
      px = px + x[i];
      py = py + y[i];
    }
  }
  for (i = 1; i <= n; i++) cout << " " << nho[i];
}
int main() {
  nhap();
  check();
  cout << "" << endl;
  return 0;
}