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 
import random

n = int(input())
v = []
a = []
for i in range(n):
    a.append(i)

for _ in range(0, n):
    x, y = map(int, input().split())
    v.append([x, y, x*x+y*y])

while 1>0:
    x = 0
    y = 0
    ans = [0]*n
    random.shuffle(a)
    for i in range(n):
        if (x+v[a[i]][0])**2+(y+v[a[i]][1])**2 <= (x-v[a[i]][0])**2+(y-v[a[i]][1])**2:
            x += v[a[i]][0]
            y += v[a[i]][1]
            ans[a[i]] = 1
        else:
            x -= v[a[i]][0]
            y -= v[a[i]][1]
            ans[a[i]] = -1
    if x*x+y*y <= 1500000**2:
        print(*ans)
        break

