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>
bool neg;
template <class T>
inline void in(T &x) {
  x = 0;
  register char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') neg = true;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - 48;
  if (neg) neg = false, x = -x;
}
struct INFO {
  long long a, b;
  int id;
} dat[100005];
int ans[100005];
inline long long calc(long long a, long long b) { return a * a + b * b; }
int main(void) {
  long long a, b;
  srand(time(NULL));
  int n;
  in(n);
  long long x, y;
  long long bd = 1500000ll * 1500000ll;
  for (register int i = 1; i <= n; ++i)
    in(dat[i].a), in(dat[i].b), dat[i].id = i;
  while (233) {
    x = y = 0;
    for (register int i = 1; i <= n; ++i)
      if (calc(x + dat[i].a, y + dat[i].b) < calc(x - dat[i].a, y - dat[i].b))
        x += dat[i].a, y += dat[i].b, ans[dat[i].id] = 1;
      else
        x -= dat[i].a, y -= dat[i].b, ans[dat[i].id] = -1;
    if (calc(x, y) <= bd) {
      for (register int i = 1; i <= n; ++i) printf("%d ", ans[i]);
      return 0;
    }
    std::random_shuffle(dat + 1, dat + 1 + n);
  }
}
