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 INF = 0x3F3F3F3F;
const int MOD = 1e9 + 7;
const int MAX = 1e5 + 5;
int N;
pair<int, int> arr[2 * MAX];
vector<pair<int, int> > adj[2 * MAX];
vector<int> v = {1, 2};
int curr;
bool vis[2 * MAX];
int dist[2 * MAX];
inline long long int distSq(pair<int, int> v) {
  return (long long int)v.first * v.first + (long long int)v.second * v.second;
}
inline pair<int, int> add(pair<int, int> a, pair<int, int> b) {
  return {a.first + b.first, a.second + b.second};
}
inline pair<int, int> sub(pair<int, int> a, pair<int, int> b) {
  return {a.first - b.first, a.second - b.second};
}
inline bool check(pair<int, int> a, pair<int, int> b, bool plus, int lim) {
  if (!plus) {
    b.first = -b.first, b.second = -b.second;
  }
  return distSq(add(a, b)) <= (long long int)lim * lim;
}
inline void findVecs(int lim) {
  for (int i = 0; i < v.size(); i++) {
    for (int j = i + 1; j < v.size(); j++) {
      if (check(arr[v[i]], arr[v[j]], 0, lim)) {
        arr[curr] = sub(arr[v[i]], arr[v[j]]);
        adj[curr].push_back({v[i], 1}), adj[curr].push_back({v[j], -1});
        v.erase(v.begin() + j), v.erase(v.begin() + i);
        return;
      } else if (check(arr[v[i]], arr[v[j]], 1, lim)) {
        arr[curr] = add(arr[v[i]], arr[v[j]]);
        adj[curr].push_back({v[i], 1}), adj[curr].push_back({v[j], 1});
        v.erase(v.begin() + j), v.erase(v.begin() + i);
        return;
      }
    }
  }
}
void dfs(int u) {
  for (pair<int, int> e : adj[u]) {
    int v = e.first, w = e.second;
    if (!vis[v]) {
      vis[v] = 1;
      dist[v] = dist[u] * w;
      dfs(v);
    }
  }
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d%d", &arr[i].first, &arr[i].second);
  }
  if (N == 1) {
    printf("1 ");
    return 0;
  }
  curr = N + 1;
  for (int i = 3; i <= N; i++) {
    v.push_back(i);
    findVecs(1e6);
    v.push_back(curr);
    curr++;
  }
  findVecs(1.5e6);
  fill(dist + 1, dist + curr + 1, 1);
  vis[curr] = 1;
  dfs(curr);
  for (int i = 1; i <= N; i++) {
    printf("%d ", dist[i]);
  }
  return 0;
}
