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 x[200005], y[200005];
int sgn[200005];
int par[200005];
vector<int> g[200005];
bool check(int a, int b, long long th) {
  int icap = x[a] + x[b];
  int jcap = y[a] + y[b];
  long long z = 1LL * icap * icap + 1LL * jcap * jcap;
  return (z <= th);
}
bool checkR(int a, int b, long long th) {
  int icap = x[a] - x[b];
  int jcap = y[a] - y[b];
  long long z = 1LL * icap * icap + 1LL * jcap * jcap;
  return (z <= th);
}
void dfs(int x, int p) {
  sgn[x] *= sgn[p];
  for (auto i : g[x]) {
    if (i != p) dfs(i, x);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n;
  cin >> n;
  for (int i = 1; i < n + 1; i++) {
    cin >> x[i] >> y[i];
  }
  vector<int> v;
  for (int i = 1; i < n + 1; i++) {
    v.push_back(i);
  }
  for (int i = 0; i < 200005; i++) {
    sgn[i] = 1;
  }
  int id = n;
  while (v.size() > 2) {
    int u[3];
    u[0] = v.back();
    v.pop_back();
    u[1] = v.back();
    v.pop_back();
    u[2] = v.back();
    v.pop_back();
    bool flag = 0;
    for (int i = 0; i < 3; i++) {
      for (int j = i + 1; j < 3; j++) {
        if (check(u[i], u[j], 1e12)) {
          par[u[i]] = par[u[j]] = ++id;
          g[id].push_back(u[i]);
          g[id].push_back(u[j]);
          x[id] = x[u[i]] + x[u[j]];
          y[id] = y[u[i]] + y[u[j]];
          v.push_back(id);
          for (int k = 0; k < 3; k++) {
            if (k != i && k != j) v.push_back(u[k]);
          }
          flag = 1;
        } else if (checkR(u[i], u[j], 1e12)) {
          par[u[i]] = par[u[j]] = ++id;
          g[id].push_back(u[i]);
          g[id].push_back(u[j]);
          x[id] = x[u[i]] - x[u[j]];
          y[id] = y[u[i]] - y[u[j]];
          sgn[u[j]] = -1;
          v.push_back(id);
          for (int k = 0; k < 3; k++) {
            if (k != i && k != j) v.push_back(u[k]);
          }
          flag = 1;
        }
        if (flag == 1) break;
      }
      if (flag) break;
    }
    assert(flag);
  }
  ++id;
  g[id].push_back(v[0]);
  g[id].push_back(v[1]);
  if (check(v[0], v[1], 2.26 * 1e12) == 0) {
    sgn[v[1]] = -1;
  }
  dfs(id, 0);
  int ansx = 0, ansy = 0;
  for (int i = 1; i < n + 1; i++) {
    ansx += x[i] * sgn[i];
    ansy += y[i] * sgn[i];
    cout << sgn[i] << " ";
  }
}