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 const N = 100000;
long long const inf = 1000000ll * 1000000ll, inf2 = 1500000ll * 1500000ll;
int n;
vector<pair<int, bool> > g[N];
bool an[N];
struct V {
  int x, y, id;
  V() {}
  V(int a, int b, int i) : x(a), y(b), id(i) {}
  void sc(int i) {
    scanf("%d%d", &x, &y);
    id = i;
  }
  V operator+(V o) { return V(x + o.x, y + o.y, id); }
  V operator-() { return V(-x, -y, id); }
  long long L() { return (long long)x * x + (long long)y * y; }
} v[N];
void ad(int i, int j, bool flp) {
  g[i].push_back(make_pair(j, flp));
  g[j].push_back(make_pair(i, flp));
}
void go(int v, int p = -1, bool cl = true) {
  an[v] = cl;
  for (int i = 0; i < (int)(g[v].size()); ++i) {
    int u = g[v][i].first;
    if (u == p) continue;
    go(u, v, cl ^ g[v][i].second);
  }
}
int main() {
  scanf("%d", &n);
  if (n == 1) {
    printf("1\n");
    return 0;
  }
  for (int i = 0; i < (int)(n); ++i) v[i].sc(i);
  int t = n;
  --n;
  while (n > 1) {
    if ((v[n] + v[n - 1]).L() <= inf)
      ad(v[n].id, v[n - 1].id, false), v[n - 1] = v[n] + v[n - 1];
    else if ((v[n] + -v[n - 1]).L() <= inf)
      ad(v[n].id, v[n - 1].id, true), v[n - 1] = v[n] + -v[n - 1];
    else if ((v[n] + v[n - 2]).L() <= inf)
      ad(v[n].id, v[n - 2].id, false), v[n - 2] = v[n] + v[n - 2];
    else if ((v[n] + -v[n - 2]).L() <= inf)
      ad(v[n].id, v[n - 2].id, true), v[n - 2] = v[n] + -v[n - 2];
    else if ((v[n - 1] + v[n - 2]).L() <= inf)
      ad(v[n - 1].id, v[n - 2].id, false), v[n - 2] = v[n - 2] + v[n - 1],
                                                 swap(v[n - 1], v[n]);
    else
      ad(v[n - 1].id, v[n - 2].id, true), v[n - 2] = v[n - 2] + -v[n - 1],
                                                swap(v[n - 1], v[n]);
    --n;
  }
  go(v[0].id);
  if ((v[0] + v[1]).L() <= inf2)
    go(v[1].id);
  else
    go(v[1].id, -1, false);
  printf("%d", an[0] ? 1 : -1);
  for (int i = 1; i < (int)(t); ++i) printf(" %d", an[i] ? 1 : -1);
  printf("\n");
}