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>
inline long long read() {
  char c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  long long k = 0, kk = 1;
  if (c == '-') c = getchar(), kk = -1;
  while (c >= '0' && c <= '9') k = k * 10 + c - '0', c = getchar();
  return k * kk;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x / 10) write(x / 10);
  putchar(x % 10 + '0');
}
void writeln(long long x) {
  write(x);
  puts("");
}
using namespace std;
pair<long long, long long> inc(pair<long long, long long> x,
                               pair<long long, long long> y) {
  x.first += y.first;
  x.second += y.second;
  return x;
}
pair<long long, long long> dec(pair<long long, long long> x,
                               pair<long long, long long> y) {
  x.first -= y.first;
  x.second -= y.second;
  return x;
}
long long clc(pair<long long, long long> x) {
  return x.first * x.first + x.second * x.second;
}
long long pd(pair<long long, long long> x, pair<long long, long long> y) {
  return clc(inc(x, y)) < clc(dec(x, y));
}
long long n, f[1000010], ff[1000010];
pair<long long, long long> a[1000010], zero;
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++)
    a[i].first = read(), a[i].second = read(), f[i] = i;
  while (1) {
    random_shuffle(f + 1, f + 1 + n);
    pair<long long, long long> x = zero;
    for (long long i = 1; i <= n; i++)
      if (pd(x, a[f[i]])) {
        ff[f[i]] = 1;
        x = inc(x, a[f[i]]);
      } else {
        ff[f[i]] = -1;
        x = dec(x, a[f[i]]);
      }
    if (clc(x) <= 1500000ll * 1500000ll) {
      for (long long i = 1; i <= n; i++) write(ff[i]), putchar(' ');
      return 0;
    }
  }
}