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 double yo = 1.5 * 1e6;
struct node {
  long long x, y, id;
} s[100005];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld%lld", &s[i].x, &s[i].y);
    s[i].id = i;
  }
  int ans[100005] = {0};
  while (1) {
    random_shuffle(s + 1, s + 1 + n);
    long long sx = 0, sy = 0;
    for (int i = 1; i <= n; ++i) {
      bool f = true;
      long long x = s[i].x, y = s[i].y;
      long long tx1 = sx + x, ty1 = sy + y, tx2 = sx - x, ty2 = sy - y;
      double ts1 = tx1 * tx1 + ty1 * ty1, ts2 = tx2 * tx2 + ty2 * ty2;
      ts1 = sqrt(ts1);
      ts2 = sqrt(ts2);
      if (ts1 < ts2) {
        f = true;
        sx = tx1, sy = ty1;
      } else if (ts1 > ts2) {
        f = false;
        sx = tx2, sy = ty2;
      } else {
        if (sx >= 0 && sy >= 0) {
          f = false;
          sx = tx2, sy = ty2;
        } else {
          f = true;
          sx = tx1, sy = ty1;
        }
      }
      ans[s[i].id] = f ? 1 : -1;
    }
    if (sx * sx + sy * sy <= yo * yo) {
      for (int i = 1; i <= n; ++i) {
        printf("%d%c", ans[i], i == n ? '\n' : ' ');
      }
      break;
    }
  }
  return 0;
}