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;
long long int n, lo, hi, mid, candies, num, x[200005], y[200005], ans[200005],
    temp, indexx, test1, test2, maxcap;
long long int dis(long long int r, long long int s) { return r * r + s * s; }
long long int input() {
  cin >> n;
  long long int currx, curry;
  curry = currx = 0;
  maxcap = 1e6;
  maxcap = sqrt(12.5) * maxcap;
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i];
    test1 = test2 = 0;
    if (abs(currx + x[i]) <= maxcap && abs(curry + y[i]) <= maxcap) {
      test1 = 1;
    }
    if (abs(currx - x[i]) <= maxcap && abs(curry - y[i]) <= maxcap) {
      test2 = 1;
    }
    if ((test1 == 1 && test2 == 1) || (test2 == 0 && test1 == 0)) {
      if (dis(currx + x[i], curry + y[i]) <= dis(currx - x[i], curry - y[i])) {
        ans[i] = 1;
        currx = currx + x[i];
        curry = curry + y[i];
      } else {
        currx = currx - x[i];
        curry = curry - y[i];
        ans[i] = -1;
      }
      continue;
    }
    if (test1 == 1) {
      ans[i] = 1;
      currx = currx + x[i];
      curry = curry + y[i];
    } else {
      ans[i] = -1;
      currx = currx - x[i];
      curry = curry - y[i];
    }
  }
  for (int j = 0; j < 2; ++j) {
    for (int i = 0; i < n; ++i) {
      if (ans[i] == 1) {
        if (dis(currx - 2 * x[i], curry - 2 * y[i]) < dis(currx, curry)) {
          ans[i] = -1;
          currx = currx - 2 * x[i];
          curry = curry - 2 * y[i];
        }
      }
      if (ans[i] == -1) {
        if (dis(currx + 2 * x[i], curry + 2 * y[i]) < dis(currx, curry)) {
          ans[i] = 1;
          currx = currx + 2 * x[i];
          curry = curry + 2 * y[i];
        }
      }
    }
  }
  if (dis(currx, curry) > maxcap * maxcap) {
    for (int i = 0; i < n; ++i) {
      if (ans[i] == 1) {
        if (dis(currx - 2 * x[i], curry - 2 * y[i]) <= maxcap * maxcap) {
          ans[i] = -1;
          break;
        }
      }
      if (ans[i] == -1) {
        if (dis(currx + 2 * x[i], curry + 2 * y[i]) <= maxcap * maxcap) {
          ans[i] = 1;
          break;
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}
int main() {
  input();
  return 0;
}