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;
struct vec {
  int x, y;
  vector<pair<int, int>> sgn;
  vec() {}
  vec(int x, int y, int idx) : x(x), y(y) { sgn = {{idx, 1}}; }
  long long merge_d(const vec &v2, int mul) const {
    long long dx = x + v2.x * mul;
    long long dy = y + v2.y * mul;
    return dx * dx + dy * dy;
  }
  vec merge(const vec &v2, int mul) const {
    vec res;
    res.x = x + v2.x * mul;
    res.y = y + v2.y * mul;
    res.sgn = sgn;
    for (auto [idx, i] : v2.sgn) {
      res.sgn.push_back({idx, mul * i});
    }
    return res;
  }
};
int n;
vector<vec> pts;
long long small_limit = 1000000000000;
long long big_limit = 2250000000000;
int ans[100010];
array<vec, 2> merge(vec a, vec b, vec c) {
  if (a.merge_d(b, 1) <= small_limit) {
    return {a.merge(b, 1), c};
  } else if (a.merge_d(b, -1) <= small_limit) {
    return {a.merge(b, -1), c};
  }
  if (a.merge_d(c, 1) <= small_limit) {
    return {a.merge(c, 1), b};
  } else if (a.merge_d(c, -1) <= small_limit) {
    return {a.merge(c, -1), b};
  }
  if (b.merge_d(c, 1) <= small_limit) {
    return {b.merge(c, 1), a};
  } else {
    return {b.merge(c, -1), a};
  }
}
vector<vec> solve(int l, int r) {
  if (l == r) {
    return {pts[l]};
  } else {
    int m = (l + r) / 2;
    auto list = solve(l, m);
    auto rest = solve(m + 1, r);
    list.insert(list.end(), rest.begin(), rest.end());
    while (list.size() > 2) {
      auto a = list.back();
      list.pop_back();
      auto b = list.back();
      list.pop_back();
      auto c = list.back();
      list.pop_back();
      array<vec, 2> res = merge(a, b, c);
      list.insert(list.end(), res.begin(), res.end());
    }
    return list;
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    pts.emplace_back(x, y, i);
  }
  if (n == 1) {
    cout << "1\n";
    return 0;
  }
  vector<vec> ans_v = solve(0, n - 1);
  vec res;
  if (ans_v[0].merge_d(ans_v[1], 1) <= big_limit) {
    res = ans_v[0].merge(ans_v[1], 1);
  } else {
    res = ans_v[0].merge(ans_v[1], -1);
  }
  for (auto [idx, i] : res.sgn) {
    ans[idx] = i;
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
  return 0;
}
