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;
const long long MAXN = 1500000;
const long long maxn = 1000000 + 5;
int z[maxn];
typedef struct node {
  long long x;
  long long y;
  long long mark;
} Node;
vector<Node> v;
int main() {
  long long a = 0, b = 0, p = MAXN * MAXN;
  Node mid;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%I64d%I64d", &mid.x, &mid.y);
    mid.mark = i;
    v.push_back(mid);
  }
  for (int i = 0; i < n; i++) {
    if (a >= 0 && b >= 0 && v[i].x <= 0 && v[i].y <= 0) {
      a += v[i].x;
      b += v[i].y;
      z[v[i].mark] = 1;
    } else if (a >= 0 && b >= 0 && v[i].x >= 0 && v[i].y >= 0) {
      a -= v[i].x;
      b -= v[i].y;
      z[v[i].mark] = -1;
    } else if (a <= 0 && b <= 0 && v[i].x <= 0 && v[i].y <= 0) {
      a -= v[i].x;
      b -= v[i].y;
      z[v[i].mark] = -1;
    } else if (a <= 0 && b <= 0 && v[i].x >= 0 && v[i].y >= 0) {
      a += v[i].x;
      b += v[i].y;
      z[v[i].mark] = 1;
    } else if ((a - v[i].x) * (a - v[i].x) + (b - v[i].y) * (b - v[i].y) <
               (a + v[i].x) * (a + v[i].x) + (b + v[i].y) * (b + v[i].y)) {
      a -= v[i].x;
      b -= v[i].y;
      z[v[i].mark] = -1;
    } else {
      a += v[i].x;
      b += v[i].y;
      z[v[i].mark] = 1;
    }
    if (i == n - 1) {
      if (a * a + b * b > p) {
        random_shuffle(v.begin(), v.end());
        i = 0;
        a = 0;
        b = 0;
        continue;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", z[i]);
  }
  return 0;
}