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 = 2e5 + 5;
const int r = 1e6;
const int lim = 1.5e6;
struct point {
  int x, y;
  point() {}
  point(int _x, int _y) : x(_x), y(_y) {}
  point operator+(const point& other) const {
    return point(x + other.x, y + other.y);
  }
  point operator*(int k) const { return point(x * k, y * k); }
};
long long sqr_len(point u) { return u.x * 1LL * u.x + u.y * 1LL * u.y; }
int n;
point moves[maxn];
int nvertex;
vector<int> gr[maxn];
int state[maxn];
int ans[maxn];
void read_input() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> moves[i].x >> moves[i].y;
}
void dfs(int u, int cur) {
  cur ^= state[u];
  if (cur)
    ans[u] = -1;
  else
    ans[u] = 1;
  for (auto& v : gr[u]) dfs(v, cur);
}
void solve() {
  if (n == 1) {
    cout << "1\n";
    return;
  }
  vector<pair<point, int>> cur;
  for (int i = 1; i <= 2; ++i) cur.push_back(make_pair(moves[i], i));
  nvertex = n;
  for (int i = 3; i <= n; ++i) {
    cur.push_back(make_pair(moves[i], i));
    bool ok = false;
    ++nvertex;
    for (int j = 0; j < 3; ++j)
      for (int k = j + 1; k < 3; ++k)
        for (int signj = -1; signj < 2; signj += 2)
          for (int signk = -1; signk < 2; signk += 2)
            if (!ok && sqr_len(cur[j].first * signj + cur[k].first * signk) <=
                           r * 1LL * r) {
              ok = true;
              point new_move = cur[j].first * signj + cur[k].first * signk;
              if (signj == -1) state[cur[j].second] = 1;
              if (signk == -1) state[cur[k].second] = 1;
              gr[nvertex].push_back(cur[j].second);
              gr[nvertex].push_back(cur[k].second);
              pair<point, int> temp;
              for (int t = 0; t < 3; ++t)
                if (t != j && t != k) temp = cur[t];
              cur.clear();
              cur.push_back(temp);
              cur.push_back(make_pair(new_move, nvertex));
            }
    assert(ok);
  }
  ++nvertex;
  bool ok = false;
  for (int j = 0; j < 2; ++j)
    for (int k = j + 1; k < 2; ++k)
      for (int signj = -1; signj < 2; signj += 2)
        for (int signk = -1; signk < 2; signk += 2)
          if (!ok && sqr_len(cur[j].first * signj + cur[k].first * signk) <=
                         lim * 1LL * lim) {
            ok = true;
            if (signj == -1) state[cur[j].second] = 1;
            if (signk == -1) state[cur[k].second] = 1;
            gr[nvertex].push_back(cur[j].second);
            gr[nvertex].push_back(cur[k].second);
          }
  assert(ok);
  dfs(nvertex, 0);
  for (int i = 1; i <= n; ++i) cout << ans[i] << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  read_input();
  solve();
}