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 
n=input()
x=0
y=0
def dis(x,y,a,b):
  return (x-a)**2+(y-b)**2
lst=[]
ans=[0]*n
lst2=[]
lst4=[]
ans2=[0]*n
for i in range(0,n):
  a,b=map(int,raw_input().split())
  lst4.append((a,b))
  lst.append((a*a,a,b,i))
  lst2.append((b*b,a,b,i))
lst.sort(reverse=True)
lst2.sort(reverse=True)
for i in range(0,n):
  a=lst[i][1]
  b=lst[i][2]
  val=lst[i][3]
  if dis(a,b,x,y)<dis(a,b,-x,-y):
    ans[val]=str(-1)
    x=x-a
    y=y-b 
  else:
    ans[val]=str(1)
    x=x+a 
    y=y+b
if n<1000:
    
  for i in range(0,n):
    a=lst2[i][1]
    b=lst2[i][2]
    val=lst2[i][3]
    if dis(a,b,x,y)<dis(a,b,-x,-y):
      ans2[val]=str(-1)
      x=x-a
      y=y-b 
    else:
      ans2[val]=str(1)
      x=x+a 
      y=y+b
  s1=0
  s2=0
  s3=0
  s4=0
  for i in range(0,n):
    s1+=int(ans[i])*lst4[i][0]
    s2+=int(ans[i])*lst4[i][1]
    s3+=int(ans2[i])*lst4[i][0]
    s4+=int(ans2[i])*lst4[i][1]
  if s1**2+s2**2<s3**2+s4**2:
    print " ".join(ans)
  else:
    print " ".join(ans2)
else:
  
  print " ".join(ans)




