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;
struct vec {
  int num;
  double x, y;
  double mod() { return sqrt(x * x + y * y); }
  vec operator+(vec v) { return (vec){0, x + v.x, y + v.y}; }
  vec operator*(int t) { return (vec){0, t * x, t * y}; }
} v[100010];
int n, f[100010], c[100010], C, tag[100010];
queue<vec> q;
inline void link(int ca, vec &a, int cb, vec &b) {
  f[b.num] = a.num;
  tag[a.num] *= ca;
  tag[b.num] *= cb * tag[a.num];
  q.push((vec){a.num, ca * a.x + cb * b.x, ca * a.y + cb * b.y});
}
void init() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    v[i].num = i;
    tag[i] = 1;
    f[i] = i;
    scanf("%lf%lf", &v[i].x, &v[i].y);
    q.push(v[i]);
  }
}
void work() {
  vec v1, v2, v3;
  int c1, c2, c3, flag;
  while (q.size() > 2) {
    v1 = q.front(), q.pop();
    v2 = q.front(), q.pop();
    v3 = q.front(), q.pop();
    flag = 1;
    for (c1 = -1; c1 <= 1 and flag; c1 += 2)
      for (c2 = -1; c2 <= 1 and flag; c2 += 2)
        for (c3 = -1; c3 <= 1 and flag; c3 += 2) {
          if ((v1 * c1 + v2 * c2).mod() < (1e6 + 1e-8))
            link(c1, v1, c2, v2), q.push(v3), flag = 0;
          else if ((v1 * c1 + v3 * c3).mod() < (1e6 + 1e-8))
            link(c1, v1, c3, v3), q.push(v2), flag = 0;
          else if ((v2 * c2 + v3 * c3).mod() < (1e6 + 1e-8))
            link(c2, v2, c3, v3), q.push(v1), flag = 0;
        }
  }
  v1 = q.front(), q.pop();
  v2 = q.front(), q.pop();
  if ((v1 + v2).mod() < (v1 + v2 * (-1)).mod())
    link(1, v1, 1, v2);
  else
    link(1, v1, -1, v2);
  v1 = q.front();
}
int calc(int x) {
  if (c[x]) return c[x];
  if (x == f[x])
    c[x] = tag[x];
  else
    c[x] = tag[x] * calc(f[x]);
  return c[x];
}
int main() {
  init();
  if (n == 1) {
    printf("1");
    return 0;
  }
  work();
  for (int i = 1; i <= n; i++)
    if (!c[i]) calc(i);
  for (int i = 1; i <= n; i++) printf("%d ", c[i]);
  return 0;
}