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;
unsigned int gcd(unsigned int u, unsigned int v) {
  int shift;
  if (u == 0) return v;
  if (v == 0) return u;
  shift = __builtin_ctz(u | v);
  u >>= __builtin_ctz(u);
  do {
    v >>= __builtin_ctz(v);
    if (u > v) {
      unsigned int t = v;
      v = u;
      u = t;
    }
    v = v - u;
  } while (v != 0);
  return u << shift;
}
bool cmp(pair<int64_t, pair<int64_t, int64_t>> a,
         pair<int64_t, pair<int64_t, int64_t>> b) {
  return abs(a.first) + abs(a.second.first) <
         abs(b.second.first) + abs(b.first);
}
int64_t dist(int64_t x, int64_t y) { return x * x + y * y; }
int64_t sq(int64_t n) { return n * n; }
int main() {
  int64_t i, j, q, k, t, m, n;
  cin >> n;
  ;
  pair<int64_t, pair<int64_t, int64_t>> v[n];
  for (i = 0; i < n; i++) {
    cin >> v[i].first;
    ;
    cin >> v[i].second.first;
    ;
    v[i].second.second = i;
  }
  int64_t x, y;
  x = 0;
  y = 0;
  sort(v, v + n, cmp);
  int64_t ans[n];
  for (i = n - 1; i >= 0; i--) {
    if (dist(x + v[i].first, y + v[i].second.first) <=
        dist(x - v[i].first, y - v[i].second.first))
      ans[v[i].second.second] = 1;
    else
      ans[v[i].second.second] = -1;
    x += (ans[v[i].second.second] * v[i].first);
    y += (ans[v[i].second.second] * v[i].second.first);
  }
  if (dist(x, y) > (sq(1500000)))
    for (i = 0; i < n; i++) ans[i] *= -1;
  for (i = 0; i < n; i++) cout << ans[i] << " ";
}