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;
const double R = 1.5e6;
int n, pos[N + 2], con, nn;
struct point {
  double x, y;
} p[N + 2];
double mag(point pp) { return sqrt(pp.x * pp.x + pp.y * pp.y); }
void dfs(int msk, int cnt, point pp) {
  if (cnt > nn) {
    point ppp = pp;
    for (int i = 1; i <= nn; i++) {
      int tp = msk & (1 << i);
      if (tp) {
        ppp.x += p[con + i].x, ppp.y += p[con + i].y;
        pos[con + i] = 1;
      } else {
        ppp.x -= p[con + i].x, ppp.y -= p[con + i].y;
        pos[con + i] = -1;
      }
    }
    if (mag(ppp) <= R) {
      for (int i = 1; i <= n; i++) cout << pos[i] << " ";
      cout << endl;
      exit(0);
    }
    return;
  }
  dfs(msk, cnt + 1, pp);
  dfs(msk | (1 << cnt), cnt + 1, pp);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i].x >> p[i].y;
  point pp = {0, 0}, tm1, tm2;
  for (int i = 1; i <= n - 3; i++) {
    tm1 = {pp.x + p[i].x, pp.y + p[i].y};
    tm2 = {pp.x - p[i].x, pp.y - p[i].y};
    if (mag(tm1) <= mag(tm2)) {
      pp = tm1;
      pos[i] = 1;
    } else {
      pp = tm2;
      pos[i] = -1;
    }
  }
  con = max(0, n - 3);
  nn = n - con;
  dfs(0, 1, pp);
  return 0;
}