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;
int n;
vector<vector<int>> v1, v2, v3;
int r[101010];
void lol(vector<vector<int>> &v) {
  for (int i = 0; i < v.size(); i++) {
    if (v[i][0] < 0) {
      v[i][0] = -v[i][0];
      v[i][1] = -v[i][1];
      v[i][3] = -v[i][3];
    }
  }
  sort(v.begin(), v.end());
  int cc = 0;
  for (int i = 0; i + 1 < v.size(); i += 2) {
    v[i + 1][0] = -v[i + 1][0];
    v[i + 1][1] = -v[i + 1][1];
    v[i + 1][3] = -v[i + 1][3];
    int uu = v[i][1] + v[i + 1][1];
    if ((cc > 0 && uu > 0) || (cc < 0 && uu < 0)) {
      v[i][0] = -v[i][0];
      v[i][1] = -v[i][1];
      v[i + 1][0] = -v[i + 1][0];
      v[i + 1][1] = -v[i + 1][1];
      v[i][3] = -v[i][3];
      v[i + 1][3] = -v[i + 1][3];
    }
    cc += v[i][1] + v[i + 1][1];
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    if (abs(x) >= 500000) {
      v1.push_back({x, y, i, 1});
    } else if (abs(y) >= 500000) {
      v2.push_back({x, y, i, 1});
    } else {
      v3.push_back({x, y, i, 1});
    }
  }
  lol(v1);
  for (int i = 0; i < v2.size(); i++) swap(v2[i][0], v2[i][1]);
  lol(v2);
  for (int i = 0; i < v2.size(); i++) swap(v2[i][0], v2[i][1]);
  lol(v3);
  for (int a = -1; a <= 1; a++)
    for (int b = -1; b <= 1; b++)
      for (int c = -1; c <= 1; c++) {
        if (a == 0 || b == 0 || c == 0) continue;
        double aa = 0, bb = 0;
        for (auto x : v1) {
          aa += a * x[0];
          bb += a * x[1];
        }
        for (auto x : v2) {
          aa += b * x[0];
          bb += b * x[1];
        }
        for (auto x : v3) {
          aa += c * x[0];
          bb += c * x[1];
        }
        if (sqrt(aa * aa + bb * bb) <= 1.5 * 1e6) {
          for (auto x : v1) r[x[2]] = a * x[3];
          for (auto x : v2) r[x[2]] = b * x[3];
          for (auto x : v3) r[x[2]] = c * x[3];
          for (int i = 0; i < n; i++) cout << r[i] << " ";
          cout << "\n";
          return 0;
        }
      }
}