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;
int N;
struct direction {
  int index;
  long long x, y, z;
  double dist;
} directions[100000 + 5];
double dist(long long x, long long y) { return sqrt(x * x + y * y); }
bool comp1(direction l, direction r) { return l.dist > r.dist; }
bool comp2(direction l, direction r) { return l.index < r.index; }
void solution1() {
  long long x = 0, y = 0;
  for (int i = 0; i < N; i++) {
    long long x1 = x + directions[i].x, y1 = y + directions[i].y;
    long long x2 = x - directions[i].x, y2 = y - directions[i].y;
    if (0) cout << dist(x1, y1) << " " << dist(x2, y2) << endl;
    if (dist(x1, y1) < dist(x2, y2)) {
      directions[i].z = 1;
      x = x1, y = y1;
    } else {
      directions[i].z = -1;
      x = x2, y = y2;
    }
    if (0) cout << x << " " << y << endl;
  }
  int i = 0;
  while (i < N && dist(x, y) > 1500000) {
    long long x1 = x + directions[i].x * directions[i].z * -2,
              y1 = y + directions[i].y * directions[i].z * -2;
    if (dist(x1, y1) < dist(x, y)) {
      directions[i].z *= -1;
      x = x1, y = y1;
    }
    i++;
  }
  for (int i = 0; i < N; i++) cout << directions[i].z << " ";
  if (0) cout << dist(x, y) << endl;
}
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> directions[i].x >> directions[i].y;
  }
  solution1();
  return 0;
}
