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 n, ans[200000];
double nowx, nowy, ansx, ansy;
double work(double x, double y) { return sqrt(x * x + y * y); }
struct node {
  double x, y;
  int id;
} a[200000];
int cmp(node a, node b) { return rand() % 2; }
int main() {
  srand(time(0));
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lf%lf", &a[i].x, &a[i].y);
    a[i].id = i;
  }
  ansx = a[1].x, ansy = a[1].y;
  ans[1] = 1;
  for (int i = 2; i <= n; i++) {
    double d1 = work(ansx + a[i].x, ansy + a[i].y),
           d2 = work(ansx - a[i].x, ansy - a[i].y);
    if (d1 > d2) {
      ansx -= a[i].x, ansy -= a[i].y;
      ans[i] = -1;
    } else {
      ansx += a[i].x, ansy += a[i].y;
      ans[i] = 1;
    }
  }
  if (work(ansx, ansy) <= 1500000) {
    for (int i = 1; i <= n; i++) {
      printf("%lld ", ans[i]);
    }
    return 0;
  }
  ans[1] = -1;
  ansx = -a[1].x, ansy = -a[1].y;
  for (int i = 2; i <= n; i++) {
    double d1 = work(ansx + a[i].x, ansy + a[i].y),
           d2 = work(ansx - a[i].x, ansy - a[i].y);
    if (d1 > d2) {
      ansx -= a[i].x, ansy -= a[i].y;
      ans[i] = -1;
    } else {
      ansx += a[i].x, ansy += a[i].y;
      ans[i] = 1;
    }
  }
  if (work(ansx, ansy) <= 1500000) {
    for (int i = 1; i <= n; i++) {
      printf("%lld ", ans[i]);
    }
    return 0;
  }
  while (1) {
    random_shuffle(a + 1, a + 1 + n);
    ansx = a[1].x, ansy = a[1].y;
    ans[a[1].id] = 1;
    for (int i = 2; i <= n; i++) {
      double d1 = work(ansx + a[i].x, ansy + a[i].y),
             d2 = work(ansx - a[i].x, ansy - a[i].y);
      if (d1 > d2) {
        ansx -= a[i].x, ansy -= a[i].y;
        ans[a[i].id] = -1;
      } else {
        ansx += a[i].x, ansy += a[i].y;
        ans[a[i].id] = 1;
      }
    }
    if (work(ansx, ansy) <= 1500000) {
      for (int i = 1; i <= n; i++) {
        printf("%lld ", ans[i]);
      }
      return 0;
    }
    ans[a[1].id] = -1;
    ansx = -a[1].x, ansy = -a[1].y;
    for (int i = 2; i <= n; i++) {
      double d1 = work(ansx + a[i].x, ansy + a[i].y),
             d2 = work(ansx - a[i].x, ansy - a[i].y);
      if (d1 > d2) {
        ansx -= a[i].x, ansy -= a[i].y;
        ans[a[i].id] = -1;
      } else {
        ansx += a[i].x, ansy += a[i].y;
        ans[a[i].id] = 1;
      }
    }
    if (work(ansx, ansy) <= 1500000) {
      for (int i = 1; i <= n; i++) {
        printf("%lld ", ans[i]);
      }
      return 0;
    }
  }
  return 0;
}