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 ln {
  long long id;
  long long x, y;
} aln[200020];
struct mmm {
  long long tgt, typ;
};
struct mr {
  vector<mmm> mgs;
  long long sf = 1, tolz = 1;
} mrginfo[200020];
long long n, ans[202020];
long long sq(long long x) { return x * x; }
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> aln[i].x >> aln[i].y;
    aln[i].id = i;
  }
  for (int i = 0; i < n - 1; i++) {
    if (i == n - 2) break;
    for (int mt = 1; mt > -3; mt -= 2) {
      if (sq(aln[i].x + aln[i + 1].x * mt) + sq(aln[i].y + aln[i + 1].y * mt) <=
          (long long)1000000000000) {
        mrginfo[aln[i + 1].id].mgs.push_back((mmm){aln[i].id, mt});
        aln[i + 1].x = aln[i].x * mt + aln[i + 1].x;
        aln[i + 1].y = aln[i].y * mt + aln[i + 1].y;
        break;
      }
      if (sq(aln[i].x + aln[i + 2].x * mt) + sq(aln[i].y + aln[i + 2].y * mt) <=
          (long long)1000000000000) {
        mrginfo[aln[i + 2].id].mgs.push_back((mmm){aln[i].id, mt});
        aln[i + 2].x = aln[i + 2].x + aln[i].x * mt;
        aln[i + 2].y = aln[i + 2].y + aln[i].y * mt;
        break;
      }
      if (sq(aln[i + 1].x + aln[i + 2].x * mt) +
              sq(aln[i + 1].y + aln[i + 2].y * mt) <=
          (long long)1000000000000) {
        mrginfo[aln[i + 2].id].mgs.push_back((mmm){aln[i + 1].id, mt});
        aln[i + 2].x = aln[i + 2].x + aln[i + 1].x * mt;
        aln[i + 2].y = aln[i + 2].y + aln[i + 1].y * mt;
        aln[i + 1] = aln[i];
        break;
      }
    }
  }
  for (int mt = 1; mt > -3; mt -= 2) {
    if (sq(aln[n - 2].x + aln[n - 1].x * mt) +
            sq(aln[n - 2].y + aln[n - 1].y * mt) <=
        (long long)2250000000000) {
      mrginfo[n - 1].sf = mt;
      break;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    ans[i] = mrginfo[i].sf;
    for (int j = 0; j < mrginfo[i].mgs.size(); j++) {
      mrginfo[mrginfo[i].mgs[j].tgt].sf = mrginfo[i].mgs[j].typ * mrginfo[i].sf;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] << ' ';
  }
  return 0;
}