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 static const N = 200200;
long long n[4], gr[4][2][N], ind[4][N];
int znak[N];
int main() {
  int nn;
  scanf("%i", &nn);
  for (int i = 0; i < nn; i++) {
    long long a, b;
    scanf("%I64d%I64d", &a, &b);
    if (a >= 0) {
      if (b > 0) {
        ind[0][n[0]] = i;
        gr[0][0][n[0]] = a;
        gr[0][1][n[0]++] = b;
      } else {
        ind[3][n[3]] = i;
        gr[3][0][n[3]] = a;
        gr[3][1][n[3]++] = b;
      }
    } else {
      if (b > 0) {
        ind[1][n[1]] = i;
        gr[1][0][n[1]] = a;
        gr[1][1][n[1]++] = b;
      } else {
        ind[2][n[2]] = i;
        gr[2][0][n[2]] = a;
        gr[2][1][n[2]++] = b;
      }
    }
  }
  for (int i = 0; i < 4; i++) {
    if (n[i])
      znak[ind[i][0]] = 1;
    else
      continue;
    long long tx = gr[i][0][0], ty = gr[i][1][0];
    for (int j = 1; j < n[i]; j++) {
      long long zbx = tx + gr[i][0][j], zby = ty + gr[i][1][j];
      long long rzx = tx - gr[i][0][j], rzy = ty - gr[i][1][j];
      if (zbx * zbx + zby * zby <= rzx * rzx + rzy * rzy) {
        tx = zbx;
        ty = zby;
        znak[ind[i][j]] = 1;
      } else {
        tx = rzx;
        ty = rzy;
        znak[ind[i][j]] = -1;
      }
    }
    gr[i][0][0] = tx;
    gr[i][1][0] = ty;
  }
  int mn = 0;
  long long minnx = 0, minny = 0;
  for (int i = 0; i < 4; i++) {
    if (n[i]) {
      minnx -= gr[i][0][0];
      minny -= gr[i][1][0];
    }
  }
  for (int i = 1; i < 16; i++) {
    long long tx = 0, ty = 0;
    for (int j = 0; j < 4; j++) {
      if (n[j]) {
        long long s = -1;
        if ((i | (1 << j)) == i) s = 1;
        tx += s * gr[j][0][0];
        ty += s * gr[j][1][0];
      }
    }
    if (tx * tx + ty * ty < minnx * minnx + minny * minny) {
      minnx = tx;
      minny = ty;
      mn = i;
    }
  }
  for (int i = 0; i < 4; i++) {
    if (n[i] && (mn | (1 << i)) > mn) {
      for (int j = 0; j < n[i]; j++) znak[ind[i][j]] *= -1;
    }
  }
  for (int i = 0; i < nn; i++) printf("%i ", znak[i]);
  return 0;
}