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 int maxn = 1e5;
int n, now;
int c[maxn * 2 + 5];
vector<int> g[maxn * 2 + 5];
pair<int, int> p[maxn * 2 + 5];
queue<int> q;
pair<int, int> operator+(pair<int, int> x, pair<int, int> y) {
  return make_pair(x.first + y.first, x.second + y.second);
}
pair<int, int> operator-(pair<int, int> x, pair<int, int> y) {
  return make_pair(x.first - y.first, x.second - y.second);
}
bool check(pair<int, int> x) {
  long long ans = 1LL * x.first * x.first + 1LL * x.second * x.second;
  return ans <= 1000000000000LL;
}
pair<int, int> cal(int a, int b) {
  if (check(p[a] + p[b])) {
    pair<int, int> ans = p[a] + p[b];
    g[now].push_back(a), g[now].push_back(b);
    return ans;
  }
  if (check(p[a] - p[b])) {
    pair<int, int> ans = p[a] - p[b];
    g[now].push_back(a), g[now].push_back(b);
    c[b] = 1;
    return ans;
  }
  return make_pair(-1, -1);
}
void dfs(int k, int now) {
  now ^= c[k];
  c[k] = now;
  for (auto x : g[k]) dfs(x, now);
}
int main() {
  scanf("%d", &n);
  if (n == 1) {
    for (int i = 1; i <= n; ++i) printf("1 ");
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    p[i] = make_pair(x, y);
    q.push(i);
  }
  now = n;
  while (q.size() >= 3) {
    int a = q.front();
    q.pop();
    int b = q.front();
    q.pop();
    int c = q.front();
    q.pop();
    ++now;
    pair<int, int> ans = cal(a, b);
    if (ans != make_pair(-1, -1))
      q.push(c);
    else {
      ans = cal(a, c);
      if (ans != make_pair(-1, -1))
        q.push(b);
      else {
        ans = cal(b, c);
        q.push(a);
      }
    }
    p[now] = ans;
    q.push(now);
  }
  int a = q.front();
  q.pop();
  int b = q.front();
  q.pop();
  ++now;
  g[now].push_back(a), g[now].push_back(b);
  if (!check(p[a] + p[b])) c[b] = 1;
  dfs(now, 0);
  for (int i = 1; i <= n; ++i)
    if (!c[i])
      printf("1 ");
    else
      printf("-1 ");
  return 0;
}