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 vec {
  long long x;
  long long y;
  int index;
  int verdict;
};
long long length(vec a) { return a.x * a.x + a.y * a.y; }
bool cmp(vec a, vec b) {
  if (length(a) < length(b)) return 0;
  if (length(a) > length(b)) return 1;
  if (a.index < b.index) return 1;
}
vec arr[100005];
int signs[100005];
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  long long sumX = 0, sumY = 0, N;
  scanf("%I64d", &N);
  for (int i = 0; i < N; ++i) {
    scanf("%I64d%I64d", &arr[i].x, &arr[i].y);
    arr[i].index = i;
  }
  while (1) {
    sumX = 0;
    sumY = 0;
    for (int i = 0; i < N; ++i) {
      vec plusSign;
      vec minusSign;
      plusSign.x = sumX + arr[i].x;
      plusSign.y = sumY + arr[i].y;
      minusSign.x = sumX - arr[i].x;
      minusSign.y = sumY - arr[i].y;
      if ((length(plusSign) < length(minusSign) && rand() % 10 < 8) ||
          (length(plusSign) >= length(minusSign) && rand() % 10 >= 8)) {
        arr[i].verdict = 1;
        sumX = sumX + arr[i].x;
        sumY = sumY + arr[i].y;
        signs[arr[i].index] = 1;
      } else {
        arr[i].verdict = -1;
        sumX = sumX - arr[i].x;
        sumY = sumY - arr[i].y;
        signs[arr[i].index] = -1;
      }
    }
    if (sqrt(sumX * sumX + sumY * sumY) < 1500000) break;
  }
  for (int i = 0; i < N; ++i) printf("%d ", signs[i]);
}