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 int MAX = 2e5 + 5, MAGIC = 20, VAL = 1500000;
int n, sgn[MAX];
long long a, b;
pair<int, int> e[MAX];
int getOpt() {
  long long best = LLONG_MAX;
  int opt = -1;
  for (int i = 1; i <= n; i++) {
    sgn[i] *= -1;
    long long na = a + 2LL * sgn[i] * e[i].first;
    long long nb = b + 2LL * sgn[i] * e[i].second;
    if (na * na + nb * nb < best) {
      best = na * na + nb * nb;
      opt = i;
    }
    sgn[i] *= -1;
  }
  return opt;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", &e[i].first, &e[i].second);
  for (int i = 1; i <= MAGIC; i++) {
    for (int j = 1; j <= n; j++) {
      if (rand() & 1)
        sgn[j] = 1;
      else
        sgn[j] = -1;
    }
    a = b = 0;
    for (int j = 1; j <= n; j++) {
      a += 1LL * sgn[j] * e[j].first;
      b += 1LL * sgn[j] * e[j].second;
    }
    while (1) {
      int opt = getOpt();
      if (opt == -1) break;
      sgn[opt] *= -1;
      a += 2LL * sgn[opt] * e[opt].first;
      b += 2LL * sgn[opt] * e[opt].second;
      if (a * a + b * b <= 1LL * VAL * VAL) {
        for (int j = 1; j <= n; j++) printf("%d ", sgn[j]);
        printf("\n");
        return 0;
      }
    }
  }
  assert(false);
  return 0;
}