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;
long long int n, ans[100005];
stack<pair<long long int, pair<long long int, long long int> > > st[5];
long long int modsq(pair<long long int, long long int> pt) {
  return pt.first * pt.first + pt.second * pt.second;
}
long long int whereto(long long int x, long long int y) {
  if (y < 0) return whereto(-x, -y);
  if (x >= 0 && x >= y)
    return 1;
  else if (x >= 0)
    return 2;
  else if (x < 0 && (-x) > y)
    return 4;
  return 3;
}
long long int dot(pair<long long int, long long int> p1,
                  pair<long long int, long long int> p2) {
  return p1.first * p2.first + p1.second * p2.second;
}
vector<long long int> g[100005];
long long int neg[100005];
void dfs(long long int st, long long int ct) {
  ct += neg[st];
  if (ct % 2) ans[st] = -1;
  for (long long int v : g[st]) dfs(v, ct);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (long long int i = 1; i <= n; ++i) ans[i] = 1;
  for (long long int i = 1; i <= n; ++i) {
    long long int x, y;
    cin >> x >> y;
    st[whereto(x, y)].push(make_pair(i, make_pair(x, y)));
  }
  while (1) {
    long long int ff = 0;
    for (long long int i = 1; i <= 4; ++i)
      if (st[i].size() > 1) ff = 1;
    if (!ff) break;
    for (long long int i = 1; i <= 4; ++i)
      if (st[i].size() > 1) {
        auto xp1 = st[i].top();
        st[i].pop();
        auto xp2 = st[i].top();
        st[i].pop();
        pair<long long int, long long int> pt1 = xp1.second, pt2 = xp2.second;
        if (dot(pt1, pt2) > 0) {
          pt2 = make_pair(-pt2.first, -pt2.second);
          neg[xp2.first] = 1;
        }
        g[xp1.first].push_back(xp2.first);
        xp1.second.first = pt1.first + pt2.first;
        xp1.second.second = pt1.second + pt2.second;
        assert(modsq(xp1.second) <= 1000000000000LL);
        st[whereto(xp1.second.first, xp1.second.second)].push(xp1);
      }
  }
  vector<pair<long long int, pair<long long int, long long int> > > tem;
  for (long long int i = 1; i <= 4; ++i)
    if (st[i].size()) tem.push_back(st[i].top());
  if (tem.size() > 1) {
    for (long long int i = 1; i <= tem.size() - 1; ++i)
      g[tem[0].first].push_back(tem[i].first);
    for (long long int mask = 0; mask <= (1LL << (tem.size() - 1)) - 1;
         ++mask) {
      pair<long long int, long long int> cur = tem[0].second;
      for (long long int i = 1; i <= tem.size() - 1; ++i) {
        if (mask & (1LL << (i - 1))) {
          neg[tem[i].first] = 1;
          cur.first -= tem[i].second.first;
          cur.second -= tem[i].second.second;
        } else {
          neg[tem[i].first] = 0;
          cur.first += tem[i].second.first;
          cur.second += tem[i].second.second;
        }
      }
      if (modsq(cur) <= 2250000000000LL) break;
    }
  }
  dfs(tem[0].first, 0);
  for (long long int i = 1; i <= n; ++i) cout << ans[i] << " ";
  cout << '\n';
}
