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;
template <class t>
inline t read(t &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t, class... A>
inline void read(t &x, A &...a) {
  read(x);
  read(a...);
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const long long N = 1e5 + 5;
long long n, a[N], b[N], id[N], prex[N], prey[N], sufx[N], sufy[N];
bool tg[N];
bool check(long long first, long long second) {
  return sqrt(first * first + second * second) <= 1500000;
}
signed main() {
  read(n);
  for (long long i = 1; i <= n; i++) read(a[i], b[i]), id[i] = i;
  for (;;) {
    random_shuffle(id + 1, id + 1 + n);
    for (long long i = 1; i <= n; i++)
      prex[i] = prex[i - 1] + a[id[i]], prey[i] = prey[i - 1] + b[id[i]];
    for (long long i = n; i; i--)
      sufx[i] = sufx[i + 1] + a[id[i]], sufy[i] = sufy[i + 1] + b[id[i]];
    for (long long i = 0; i <= n; i++)
      if (check(prex[i] - sufx[i + 1], prey[i] - sufy[i + 1])) {
        for (long long j = i + 1; j <= n; j++) tg[id[j]] = 1;
        for (long long i = 1; i <= n; i++) write(tg[i] ? -1 : 1), putchar(' ');
        return 0;
      }
  }
}