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;
int Gcd(int a, int b) {
  if (b == 0) return a;
  return Gcd(b, a % b);
}
int Lcm(int a, int b) { return a / Gcd(a, b) * b; }
inline long long read() {
  long long f = 1, x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int maxn = 1e6 + 10;
const long long inf = 1500000;
pair<pair<int, int>, long long> p[maxn];
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) {
    p[i].first.first = read();
    p[i].first.second = read();
    p[i].second = i;
  }
  while (1) {
    long long nx = 0, ny = 0, res = 0;
    vector<int> ans(n + 1, 1);
    for (int i = 1; i <= n; i++) {
      long long tx = nx - p[i].first.first, ty = ny - p[i].first.second;
      long long px = nx + p[i].first.first, py = ny + p[i].first.second;
      if (tx * tx + ty * ty < px * px + py * py) {
        res = tx * tx + ty * ty;
        nx = tx;
        ny = ty;
        ans[p[i].second] = -1;
      } else {
        res = px * px + py * py;
        nx = px;
        ny = py;
        ans[p[i].second] = 1;
      }
    }
    if (res <= inf * inf) {
      for (int i = 1; i <= n; i++) {
        printf("%d ", ans[i]);
      }
      printf("\n");
      break;
    } else {
      random_shuffle(p + 1, p + n + 1);
    }
  }
  return 0;
}
