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 long long MAXL = 1e12;
const long long CANL = 2250000000000;
struct PT {
  long long x, y;
  PT operator+(const PT& p) const { return {x + p.x, y + p.y}; }
  PT operator*(const long long delta) const { return {x * delta, y * delta}; }
  long long length() { return x * x + y * y; }
};
const int N = 3e5;
PT p[N];
pair<int, int> go[N];
int used[N], ans[N];
void dfs(int v) {
  used[v] = 1;
  int u = go[v].first, sgn = go[v].second;
  if (u == -1) return;
  if (!used[u]) dfs(u);
  ans[v] = sgn * ans[u];
}
void solve() {
  int n;
  cin >> n;
  vector<int> ids(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i].x >> p[i].y;
    ids[i] = i;
  }
  fill(go, go + 3 * n, make_pair(-1, -1));
  fill(ans, ans + 3 * n, 1);
  int timer = n;
  while (ids.size() > 2) {
    int id1 = ids.back();
    ids.pop_back();
    int id2 = ids.back();
    ids.pop_back();
    int id3 = ids.back();
    ids.pop_back();
    vector<pair<int, int> > candidates = {{id1, id2}, {id1, id3}, {id2, id3}};
    for (auto [v, u] : candidates) {
      int f = 0;
      for (int sgn = -1; sgn <= 1 && !f; sgn += 2) {
        PT np = p[v] + p[u] * sgn;
        if (np.length() <= MAXL) {
          go[v] = {timer, 1};
          go[u] = {timer, sgn};
          p[timer] = np;
          ids.push_back(timer);
          timer++;
          for (int w : {id1, id2, id3})
            if (w != v && w != u) ids.push_back(w);
          f = 1;
        }
      }
      if (f) break;
    }
  }
  if (ids.size() == 2) {
    int id1 = ids.back();
    ids.pop_back();
    int id2 = ids.back();
    ids.pop_back();
    for (int sgn = -1; sgn <= 1; sgn += 2) {
      PT np = p[id1] + p[id2] * sgn;
      if (np.length() <= CANL) {
        go[id1] = {timer, 1};
        go[id2] = {timer, sgn};
        timer++;
        break;
      }
    }
  }
  for (int v = 0; v < n; v++) {
    if (!used[v]) {
      dfs(v);
    }
    cout << ans[v] << " ";
  }
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  solve();
}