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;
struct pt {
  long long x, y;
  pt() : x(), y() {}
  pt(long long _x, long long _y) : x(_x), y(_y){};
  pt operator+(const pt &p) const { return pt(x + p.x, y + p.y); }
  pt operator-(const pt &p) const { return pt(x - p.x, y - p.y); }
  pt operator*(const long long k) const { return pt(x * k, y * k); }
  long long sqr_len() const { return x * x + y * y; }
  void in() { cin >> x >> y; }
};
long long N = (long long)1e6 * (long long)1e6;
pt a[1 << 20];
pair<pt, int> cur[4];
queue<pair<pt, int> > q;
int sig[1 << 22];
int n;
vector<int> g[1 << 22];
void dfs(int u, int i) {
  for (int v : g[u]) {
    sig[v] *= i;
    dfs(v, sig[v]);
  }
}
long long s[2] = {1, -1};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) a[i].in();
  if (n == 1) {
    cout << "1\n";
    return 0;
  }
  for (int i = 0; i < n; i++) q.push(make_pair(a[i], i));
  int m = n;
  while (q.size() > 2) {
    for (int i = 0; i < 3; i++) {
      cur[i] = q.front();
      q.pop();
    }
    bool f = 0;
    for (int s1 = 0; s1 < 2 && !f; s1++)
      for (int s2 = 0; s2 < 2 && !f; s2++) {
        for (int i = 0; i < 3 && !f; i++)
          for (int j = 0; j < 3; j++) {
            if (i == j) continue;
            if (((cur[i].first * s[s1]) + (cur[j].first * s[s2])).sqr_len() <=
                N) {
              q.push(cur[i ^ j ^ 3]);
              sig[cur[i].second] = s[s1];
              sig[cur[j].second] = s[s2];
              g[m].push_back(cur[i].second);
              g[m].push_back(cur[j].second);
              q.push(make_pair(
                  ((cur[i].first * s[s1]) + (cur[j].first * s[s2])), m++));
              f = 1;
              break;
            }
          }
      }
  }
  for (int i = 0; i < 2; i++) cur[i] = q.front(), q.pop();
  bool f = 0;
  for (int i = 0; i < 2 && !f; i++)
    for (int j = 0; j < 2; j++) {
      if (i == j) continue;
      if ((cur[i].first + cur[j].first).sqr_len() * 2 <= 3 * N) {
        sig[cur[i].second] = 1;
        sig[cur[j].second] = 1;
        g[m].push_back(cur[i].second);
        g[m].push_back(cur[j].second);
        f = 1;
        break;
      } else if ((cur[i].first - cur[j].first).sqr_len() * 2 <= 3 * N) {
        sig[cur[i].second] = 1;
        sig[cur[j].second] = -1;
        g[m].push_back(cur[i].second);
        g[m].push_back(cur[j].second);
        f = 1;
        break;
      } else if (((cur[i].first * -1) + cur[j].first).sqr_len() <= 3 * N) {
        sig[cur[i].second] = -1;
        sig[cur[j].second] = 1;
        g[m].push_back(cur[i].second);
        g[m].push_back(cur[j].second);
        f = 1;
        break;
      } else if (((cur[i].first * -1) - cur[j].first).sqr_len() * 2 <= 3 * N) {
        sig[cur[i].second] = -1;
        sig[cur[j].second] = -1;
        g[m].push_back(cur[i].second);
        g[m].push_back(cur[j].second);
        f = 1;
        break;
      }
    }
  dfs(m, 1);
  for (int i = 0; i < n; i++) cout << sig[i] << " \n"[i == n - 1];
}