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 MAXN = 1e5 + 5;
inline long long dist(long long x, long long y) {
  return 1ll * x * x + 1ll * y * y;
}
long long n;
pair<pair<long long, long long>, long long> a[MAXN];
long long res[MAXN];
long long xor_delta[MAXN];
long long ans[MAXN];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].second = i;
  }
  while (1) {
    mt19937 rng(std::chrono::system_clock::now().time_since_epoch().count());
    for (long long i = n; i >= 1; i--) {
      uniform_int_distribution<long long> r(1, i);
      swap(a[i], a[r(rng)]);
    }
    memset(xor_delta, 0, sizeof xor_delta);
    memset(res, 0, sizeof res);
    long long x = 0, y = 0;
    for (long long i = 1; i <= n; i++) {
      long long d1 = dist(x + a[i].first.first, y + a[i].first.second);
      long long d2 = dist(x - a[i].first.first, y - a[i].first.second);
      long long d3 = dist(-x + a[i].first.first, -y + a[i].first.second);
      long long d4 = dist(-x - a[i].first.first, -y - a[i].first.second);
      long long mn = min({d1, d2, d3, d4});
      if (d1 == mn) {
        x += a[i].first.first;
        y += a[i].first.second;
        res[i] = 1;
      } else if (d2 == mn) {
        x -= a[i].first.first;
        y -= a[i].first.second;
        res[i] = -1;
      } else if (d3 == mn) {
        xor_delta[1] ^= 1;
        xor_delta[i] ^= 1;
        x = -x;
        y = -y;
        x += a[i].first.first;
        y += a[i].first.second;
        res[i] = 1;
      } else {
        xor_delta[1] ^= 1;
        xor_delta[i] ^= 1;
        x = -x;
        y = -y;
        x -= a[i].first.first;
        y -= a[i].first.second;
        res[i] = -1;
      }
    }
    if (dist(x, y) > 2250000000000ll) {
      continue;
    }
    long long cur = 0;
    for (long long i = 1; i <= n; i++) {
      cur ^= xor_delta[i];
      if (cur) {
        res[i] = -res[i];
      }
    }
    x = 0, y = 0;
    for (long long i = 1; i <= n; i++) {
      x += res[i] * a[i].first.first;
      y += res[i] * a[i].first.second;
      ans[a[i].second] = res[i];
    }
    for (long long i = 1; i <= n; i++) {
      cout << ans[i] << ' ';
    }
    cout << endl;
    return 0;
  }
  return 0;
}