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;
int n, vect[3][100010], ans[100010];
void print() {
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  exit(0);
}
void fs(double x, double y, int k) {
  if (sqrt(x * x + y * y) > 1500000.0) return;
  if (k == n + 1) {
    print();
  }
  double tx1 = x + vect[1][k], ty1 = y + vect[2][k], tx2 = x - vect[1][k],
         ty2 = y - vect[2][k];
  if (sqrt(tx1 * tx1 + ty1 * ty1) < sqrt(tx2 * tx2 + ty2 * ty2)) {
    ans[k] = 1;
    fs(tx1, ty1, k + 1);
    ans[k] = -1;
    fs(tx2, ty2, k + 1);
  } else {
    ans[k] = -1;
    fs(tx2, ty2, k + 1);
    ans[k] = 1;
    fs(tx1, ty1, k + 1);
  }
}
double x, y, tx, ty, t1, t2;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &vect[1][i], &vect[2][i]);
    tx = x + vect[1][i];
    ty = y + vect[2][i];
    t1 = sqrt(tx * tx + ty * ty);
    tx = x - vect[1][i];
    ty = y - vect[2][i];
    t2 = sqrt(tx * tx + ty * ty);
    if (t1 > t2) {
      x -= vect[1][i];
      y -= vect[2][i];
      ans[i] = -1;
    } else {
      x += vect[1][i];
      y += vect[2][i];
      ans[i] = 1;
    }
  }
  if (sqrt(x * x + y * y) < 1500000) {
    print();
  }
  fs(0.0, 0.0, 1);
  print();
}