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 N = 2e5 + 15;
const long long maxi = (long long)1e12, M = 2.25 * 1e12;
int ch[N][2], ret[N];
pair<long long, long long> v[N];
long long sq(long long a) { return a * a; }
long long dist(pair<long long, long long> a, pair<long long, long long> b,
               int i, int j) {
  long long R = sq(a.first * i + b.first * j) + sq(a.second * i + b.second * j);
  return R;
}
void dfs(int root) {
  if (ch[root][0] == -1) {
    return;
  }
  for (int i = 0; i < 2; ++i) {
    int a = ch[root][i];
    ret[a] *= ret[root];
    dfs(a);
  }
}
int main() {
  int n;
  long long x, y;
  scanf("%d", &n);
  if (n == 1) {
    printf("1 ");
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    scanf("%lld%lld", &v[i].first, &v[i].second);
    ch[i][0] = ch[i][1] = -1;
    ret[i] = 1;
  }
  int pos = n;
  int id[3], g = 0;
  for (int i = 0; i < n; ++i) {
    id[g++] = i;
    if (g == 3) {
      bool is = 0;
      int rem;
      for (int a = 0; a < 3 && !is; ++a) {
        for (int b = a + 1; b < 3; ++b) {
          long long D = dist(v[id[a]], v[id[b]], 1, -1);
          int q = 1, w = -1;
          if (D > 1e12) D = dist(v[id[a]], v[id[b]], 1, 1), w = 1;
          if (D <= maxi) {
            is = 1;
            ch[pos][0] = id[a];
            ch[pos][1] = id[b];
            ret[pos] = 1;
            ret[id[b]] *= w;
            v[pos] = make_pair(v[id[a]].first + v[id[b]].first * w,
                               v[id[a]].second + v[id[b]].second * w);
            id[0] = id[3 - a - b];
            id[1] = pos++;
            g = 2;
            break;
          }
        }
      }
    }
  }
  int a = id[0], b = id[1];
  long long D = dist(v[a], v[b], 1, 1);
  int w = 1;
  if (D > M) {
    w = -1;
  }
  ret[pos] = 1;
  ch[pos][0] = a, ch[pos][1] = b;
  ret[b] *= w;
  dfs(pos);
  for (int i = 0; i < n; ++i) printf("%d ", ret[i]);
  cout << endl;
  return 0;
}