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 
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, mod = 1e9 + 7;
const double lim = (1e6) + (1e-9);
struct vec {
  int x, y;
} v[N];
vec operator+(vec a, vec b) {
  vec c = a;
  c.x += b.x;
  c.y += b.y;
  return c;
}
vec operator-(vec a, vec b) {
  vec c = a;
  c.x -= b.x;
  c.y -= b.y;
  return c;
}
double len(vec a) { return sqrt(1ll * a.x * a.x + 1ll * a.y * a.y); }
int ans[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> v[i].x >> v[i].y;
    ans[i] = 1;
  }
  if (n == 1) return cout << 1, 0;
  vec cur[2] = {v[0], v[1]};
  vector<int> id[2];
  id[0].push_back(0);
  id[1].push_back(1);
  for (int i = 2; i < n; i++) {
    if (len(cur[0] + v[i]) < lim) {
      id[0].push_back(i);
      cur[0] = cur[0] + v[i];
    } else if (len(cur[0] - v[i]) < lim) {
      ans[i] = -1;
      cur[0] = cur[0] - v[i];
      id[0].push_back(i);
    } else if (len(cur[1] + v[i]) < lim) {
      id[1].push_back(i);
      cur[1] = cur[1] + v[i];
    } else if (len(cur[1] - v[i]) < lim) {
      ans[i] = -1;
      id[1].push_back(i);
      cur[1] = cur[1] - v[i];
    } else if (len(cur[0] + cur[1]) < lim) {
      cur[0] = cur[0] + cur[1];
      cur[1] = v[i];
      for (int x : id[1]) id[0].push_back(x);
      id[1].clear();
      id[1].push_back(i);
    } else {
      cur[0] = cur[0] - cur[1];
      cur[1] = v[i];
      for (int x : id[1]) {
        id[0].push_back(x);
        ans[x] *= -1;
      }
      id[1].clear();
      id[1].push_back(i);
    }
  }
  if (len(cur[0] + cur[1]) > len(cur[0] - cur[1])) {
    for (int i : id[1]) ans[i] *= -1;
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  return 0;
}
