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 
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
int a[maxn] = {0};
set<int> s;
queue<int> q;
struct node {
  int x, y;
} v[maxn];
int f[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &v[i].x, &v[i].y);
  }
  double nowx = 0, nowy = 0;
  for (int i = 3; i < n; i++) {
    double nowx1 = nowx + v[i].x;
    double nowy1 = nowy + v[i].y;
    double s1 = sqrt(nowx1 * nowx1 + nowy1 * nowy1);
    double nowx2 = nowx - v[i].x;
    double nowy2 = nowy - v[i].y;
    double s2 = sqrt(nowx2 * nowx2 + nowy2 * nowy2);
    if (s1 <= s2) {
      nowx = nowx1;
      nowy = nowy1;
      f[i] = 1;
    } else {
      f[i] = -1;
      nowx = nowx2;
      nowy = nowy2;
    }
  }
  for (int i = 0; i < 7; i++) {
    int w = i;
    f[0] = w % 2;
    w >>= 1;
    f[1] = w % 2;
    w >>= 1;
    f[2] = w % 2;
    w >>= 1;
    if (f[0] == 0) f[0] = -1;
    if (f[1] == 0) f[1] = -1;
    if (f[2] == 0) f[2] = -1;
    double nowx1 = 0, nowy1 = 0;
    nowx1 += (f[0] * v[0].x + f[1] * v[1].x + f[2] * v[2].x + nowx);
    nowy1 += (f[0] * v[0].y + f[1] * v[1].y + f[2] * v[2].y + nowy);
    double s = sqrt(nowx1 * nowx1 + nowy1 * nowy1);
    if (s <= 1500000) {
      break;
    }
  }
  for (int i = 0; i < n; i++) {
    printf("%d\n", f[i]);
  }
  printf("\n");
  return 0;
}
