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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
using std::random_shuffle;
using std::sort;
const long long N = 1e6 + 5;
long long n, ans;
long long x[N], y[N];
long long p[N], res[N], best[N];
long long ansx, ansy;
inline bool cmp(long long a, long long b) {
  return ((x[a]) * (x[a])) + ((y[a]) * (y[a])) >
         ((x[b]) * (x[b])) + ((y[b]) * (y[b]));
}
signed main() {
  long long T = 30;
  register long long i;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) scanf("%lld%lld", &x[i], &y[i]);
  for (i = 1; i <= n; i++) p[i] = i;
  sort(p + 1, p + n + 1, cmp);
  for (i = 1; i <= n; i++) {
    if (((ansx + x[p[i]]) * (ansx + x[p[i]])) +
            ((ansy + y[p[i]]) * (ansy + y[p[i]])) <
        ((ansx - x[p[i]]) * (ansx - x[p[i]])) +
            ((ansy - y[p[i]]) * (ansy - y[p[i]])))
      ansx += x[p[i]], ansy += y[p[i]], res[p[i]] = 1;
    else
      ansx -= x[p[i]], ansy -= y[p[i]], res[p[i]] = -1;
  }
  ans = ((ansx) * (ansx)) + ((ansy) * (ansy));
  memmove(best + 1, res + 1, sizeof(long long) * n);
  while (T--) {
    ansx = 0;
    ansy = 0;
    random_shuffle(p + 1, p + n + 1);
    for (i = 1; i <= n; i++) {
      if (((ansx + x[p[i]]) * (ansx + x[p[i]])) +
              ((ansy + y[p[i]]) * (ansy + y[p[i]])) <
          ((ansx - x[p[i]]) * (ansx - x[p[i]])) +
              ((ansy - y[p[i]]) * (ansy - y[p[i]])))
        ansx += x[p[i]], ansy += y[p[i]], res[p[i]] = 1;
      else
        ansx -= x[p[i]], ansy -= y[p[i]], res[p[i]] = -1;
    }
    if (((ansx) * (ansx)) + ((ansy) * (ansy)) < ans) {
      ans = ((ansx) * (ansx)) + ((ansy) * (ansy));
      memmove(best + 1, res + 1, sizeof(long long) * n);
    }
  }
  for (i = 1; i <= n; i++) printf("%lld ", best[i]);
  putchar('\n');
  return 0;
}