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 
from sys import stdin, stdout

n = input()
inp = stdin.readlines()

if n == 1:
    print 1
    exit()

a = [0 for i in xrange(3 * n)]

child = [[-1, -1] for i in xrange(len(a))]
sign = [[1, 1] for i in xrange(len(a))]

i = 0
for line in inp:
    a[i] = tuple(map(int, line.split()))
    i += 1

r = 1000000
R = r * r

idx = n
free = [i for i in xrange(n)]

while len(free) >= 3:
    pos = []
    for it in xrange(3): pos.append(free.pop())

    found = 0
    for i in xrange(3):
        for j in xrange(i + 1, 3):
            # |v[i] + v[j]| <= r 
            pt = tuple([a[pos[i]][0] + a[pos[j]][0], a[pos[i]][1] + a[pos[j]][1]])

            if pt[0] * pt[0] + pt[1] * pt[1] <= R:
                child[idx][0] = pos[i]
                child[idx][1] = pos[j]

                sign[idx][0] = 1
                sign[idx][1] = 1

                a[idx] = pt
                free.append(idx)

                free.append(pos[i ^ j ^ 3])

                idx += 1
                found = 1
                break

            # |v[i] - v[j]| <= r
            pt = tuple([a[pos[i]][0] - a[pos[j]][0], a[pos[i]][1] - a[pos[j]][1]])

            if pt[0] * pt[0] + pt[1] * pt[1] <= R:
                child[idx][0] = pos[i]
                child[idx][1] = pos[j]

                sign[idx][0] = 1
                sign[idx][1] = -1

                a[idx] = pt
                free.append(idx)

                free.append(pos[i ^ j ^ 3])

                idx += 1
                found = 1
                break

        if found: break

pos = []
for i in xrange(2): pos.append(free.pop())

pt = tuple([a[pos[0]][0] + a[pos[1]][0], a[pos[0]][1] + a[pos[1]][1]])

if pt[0] * pt[0] + pt[1] * pt[1] <= 2 * R:
    child[idx][0] = pos[0]
    child[idx][1] = pos[1]

    sign[idx][0] = 1
    sign[idx][1] = 1

else:
    child[idx][0] = pos[0]
    child[idx][1] = pos[1]

    sign[idx][0] = 1
    sign[idx][1] = -1

ans = [1 for i in xrange(3 * n)]

stk = [idx]
ptr = 0

while ptr < len(stk):
    u = stk[ptr]
    v1, v2 = child[u]

    if v1 != -1 and v2 != -1:
        stk.append(v1)
        stk.append(v2)

        ans[v1] = sign[u][0] * ans[u]
        ans[v2] = sign[u][1] * ans[u]

    ptr += 1

out = [str(ans[i]) for i in xrange(n)]
stdout.write(' '.join(out))