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 size_t MAX = 100005;
const double PI = acos(-1.0);
const int MOD = 1000000007;
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const int Move[][2] = {{1, 0}, {-1, 0}, {0, 1},  {0, -1},
                       {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
long long gcd(const long long& a, const long long& b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long lcm(const long long& a, const long long& b) {
  return a / gcd(a, b) * b;
}
long long sqr(const long long& x) { return x * x; }
long long Pow(long long a, int t) {
  a %= MOD;
  long long base = 1;
  while (t > 0) {
    if (t & 1) base = base * a % MOD;
    a = a * a % MOD;
    t >>= 1;
  }
  return base;
}
struct Node {
  long long x, y, id;
  Node(long long x_ = 0, long long y_ = 0) { x = x_, y = y_; }
} v[MAX];
int ans[MAX];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> v[i].x >> v[i].y, v[i].id = i;
  while (1) {
    random_shuffle(v + 1, v + 1 + n);
    long long x = 0, y = 0;
    for (int i = 1; i <= n; i++) {
      long long tx1 = x + v[i].x, ty1 = y + v[i].y;
      long long tx2 = x - v[i].x, ty2 = y - v[i].y;
      if (sqr(tx1) + sqr(ty1) <= sqr(tx2) + sqr(ty2)) {
        x = tx1, y = ty1;
        ans[v[i].id] = 1;
      } else {
        x = tx2, y = ty2;
        ans[v[i].id] = -1;
      }
    }
    if (sqr(x) + sqr(y) <= sqr(1500000ll)) break;
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " \n"[i == n];
  }
}