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 PYTHON3):

Solution:
from math import sqrt
import sys
bound = 1.5 * 10 ** 6

def get_dist(pos):
    x,y = pos
    return sqrt(x**2 + y**2)    





l_3 = [(1,1,1),(1,1,-1),(1,-1,1),(1,-1,-1)]
l_2 = [(1,1),(1,-1)]
def solve_2(v1,v2):
    signs = None
    best = float('inf')
    return_pos = None
    for a,b in l_2:
        x = v1[0] * a + v2[0] * b 
        y = v1[1] * a + v2[1] * b 
        new_pos = (x,y)
        dist = get_dist(new_pos)
        if dist < best:
            signs = [a,b]
            best = dist
            return_pos = new_pos
    return signs, return_pos
def solve_3(v1,v2,v3):
    signs = None
    best = float('inf')
    return_pos = None
    for a,b,c in l_3:
        x = v1[0] * a + v2[0] * b + v3[0] * c
        y = v1[1] * a + v2[1] * b + v3[1] * c
        new_pos = (x,y)
        dist = get_dist(new_pos)
        if dist < best:
            signs = [a,b,c]
            best = dist
            return_pos = new_pos
    return signs, return_pos
n = int(input())
v_list = [list(map(int,input().split())) for _ in range(n)]
result = []

if n == 1:
    print (1)
    sys.exit()
if n == 2:
    s,current_pos = solve_2(v_list[0],v_list[1])
    print (*s)
    sys.exit()

s, current_pos = solve_3(v_list[0],v_list[1],v_list[2])
result += s
i = 3
while i<n:
    count = min(2,n-i)
    if count == 2:
        s,current_pos = solve_3(current_pos,v_list[i],v_list[i+1])
        result += s[1:]
        i += 2
    elif count == 1:
        s,current_pos = solve_2(current_pos,v_list[i])
        result += s[1:]
        i += 1
print (*result)