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 M = 1000000000000;
const int N = 200000 + 10;
struct re {
  long long x, y;
  re operator+(re a) { return (re){x + a.x, y + a.y}; }
  re operator-(re a) { return (re){x - a.x, y - a.y}; }
};
long long dist(re a) { return a.x * a.x + a.y * a.y; }
struct result {
  int cs, x, y;
};
bool check(re a) { return dist(a) <= M; }
int n, cur, f[N];
re a[N];
vector<result> res;
long long Abs(long long x) { return (x > 0) ? x : -x; }
int main() {
  cin >> n;
  for (auto i = (1); (i <= (n)); ++i) cin >> a[i].x >> a[i].y;
  if (n == 1)
    cout << 1;
  else {
    set<int> b;
    vector<int> p;
    cur = n;
    for (auto i = (1); (i <= (n)); ++i) {
      b.insert(i);
      if (b.size() == 3) {
        p.clear();
        p.insert(p.begin(), b.begin(), b.end());
        cur++;
        if (check(a[p[0]] + a[p[1]]))
          res.push_back((result){cur, p[0], p[1]}), b.erase(p[0]),
              b.erase(p[1]);
        else if (check(a[p[0]] - a[p[1]]))
          res.push_back((result){cur, p[0], -p[1]}), b.erase(p[0]),
              b.erase(p[1]);
        else if (check(a[p[0]] + a[p[2]]))
          res.push_back((result){cur, p[0], p[2]}), b.erase(p[0]),
              b.erase(p[2]);
        else if (check(a[p[0]] - a[p[2]]))
          res.push_back((result){cur, p[0], -p[2]}), b.erase(p[0]),
              b.erase(p[2]);
        else if (check(a[p[1]] + a[p[2]]))
          res.push_back((result){cur, p[1], p[2]}), b.erase(p[1]),
              b.erase(p[2]);
        else if (check(a[p[1]] - a[p[2]]))
          res.push_back((result){cur, p[1], -p[2]}), b.erase(p[1]),
              b.erase(p[2]);
        a[cur] = (res.back().y > 0) ? a[res.back().x] + a[res.back().y]
                                    : a[res.back().x] - a[-res.back().y];
        b.insert(cur);
      }
    }
    p.clear();
    p.insert(p.begin(), b.begin(), b.end());
    cur++;
    if (dist(a[p[0]] + a[p[1]]) < dist(a[p[0]] - a[p[1]]))
      res.push_back((result){cur, p[0], p[1]}), b.erase(p[0]), b.erase(p[1]);
    else
      res.push_back((result){cur, p[0], -p[1]}), b.erase(p[0]), b.erase(p[1]);
    f[cur] = 1;
    for (auto i = res.rbegin(); i != res.rend(); i++) {
      auto u = *i;
      f[u.x] = f[u.cs];
      f[Abs(u.y)] = (u.y > 0) ? f[u.cs] : -f[u.cs];
    }
    for (auto i = (1); (i <= (n)); ++i) cout << f[i] << " ";
  }
}