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 
def length_squared(v):
	return v[0]**2+v[1]**2
def add(a, b):
	return (a[0]+b[0], a[1]+b[1])
def neg(a):
	return (-a[0], -a[1])
limit=1000000000000
n=input()
cnt=0
cur=[]
source=[]
value=[]
result=[]
def test(i, j):
	global cur, cnt
	if length_squared(add(cur[i][0], cur[j][0]))<=limit:
		value.append(None)
		source.append((cur[i][1], cnt, 1))
		source.append((cur[j][1], cnt, 1))
		cur=[cur[3-(i+j)], (add(cur[i][0], cur[j][0]), cnt)]
		cnt+=1
		return True
	if length_squared(add(cur[i][0], neg(cur[j][0])))<=limit:
		value.append(None)
		source.append((cur[i][1], cnt, 1))
		source.append((cur[j][1], cnt, -1))
		cur=[cur[3-(i+j)], (add(cur[i][0], neg(cur[j][0])), cnt)]
		cnt+=1
		return True
	return False
for i in xrange(n):
	v=map(int, raw_input().split())
	result+=[cnt]
	value+=[None]
	cur+=[(v, cnt)]
	cnt+=1
	if len(cur)==3:
		assert test(0, 1) or test(0, 2) or test(1, 2)
assert 0<len(cur)<=2
assert len(result)==n
assert len(value)==cnt
assert cnt==n*2-len(cur)
assert len(source)==cnt-len(cur)
if len(cur)==1:
	value[cur[0][1]]=1
else:
	value[cur[0][1]]=1
	if length_squared(add(cur[0][0], cur[1][0]))<=length_squared(add(cur[0][0], neg(cur[1][0]))):
		value[cur[1][1]]=1
	else:
		value[cur[1][1]]=-1
for a, b, coeff in reversed(source):
	assert a<b<cnt
	assert value[a]==None
	assert value[b]!=None
	value[a]=value[b]*coeff
print " ".join(map(str, [value[result[i]] for i in xrange(n)]))
