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 int maxn = 2e5 + 5;
const int mx = 1e6;
int n, cnt;
int pa[maxn], val[maxn], ans[maxn];
long long lim = 1ll * mx * mx;
struct node {
  long long x, y;
  int id;
  node(long long x, long long y) : x(x), y(y), id(0) {}
  node() {}
  node operator+(const node &other) const {
    return node(x + other.x, y + other.y);
  }
  node operator-(const node &other) const {
    return node(x - other.x, y - other.y);
  }
  long long len() { return 1ll * x * x + 1ll * y * y; }
  void print() { printf("%lld %lld\n", x, y); }
} a[maxn], p[3];
deque<node> q;
long long cal(const node &a, const node &b, int sig) {
  return sig == 1 ? (a + b).len() : (a - b).len();
}
void ins(int u, int v, int k) {
  pa[u] = v;
  val[u] = k;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d%I64d", &a[i].x, &a[i].y);
    a[i].id = i;
    q.push_back(a[i]);
  }
  if (n == 1) {
    puts("1");
    return 0;
  }
  cnt = n;
  while (q.size() >= 3) {
    bool flag = false;
    for (int i = 0; i < 3; ++i) p[i] = q.front(), q.pop_front();
    for (int i = 0; i < 3 && !flag; ++i)
      for (int j = i + 1; j < 3 && !flag; ++j)
        for (int k = -1; k <= 1; k += 2)
          if (cal(p[i], p[j], k) <= lim) {
            flag = true;
            node tmp;
            if (k > 0)
              tmp = p[i] + p[j];
            else
              tmp = p[i] - p[j];
            tmp.id = ++cnt;
            q.push_back(tmp);
            q.push_front(p[3 - i - j]);
            ins(p[i].id, tmp.id, 1);
            ins(p[j].id, tmp.id, k);
            break;
          }
  }
  p[0] = q.front();
  q.pop_front();
  p[1] = q.front();
  q.pop_front();
  for (int k = -1; k <= 1; k += 2)
    if (cal(p[0], p[1], k) <= 2 * lim) {
      node tmp;
      if (k > 0)
        tmp = p[0] + p[1];
      else
        tmp = p[0] - p[1];
      tmp.id = ++cnt;
      ins(p[0].id, tmp.id, 1);
      ins(p[1].id, tmp.id, k);
      break;
    }
  for (int i = 1; i <= n; ++i) {
    ans[i] = 1;
    int u = i;
    while (pa[u]) {
      ans[i] *= val[u];
      u = pa[u];
    }
    printf("%d ", ans[i]);
  }
  puts("");
  return 0;
}