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 
#!/usr/bin/env python3
# -*- coding: utf-8 -*-

import time

def len2(x, y):
    return x*x+y*y

def test(v1, v2):

    x1 = v1[0]+v2[0]
    y1 = v1[1]+v2[1]
    x2 = v1[0]-v2[0]
    y2 = v1[1]-v2[1]

    l1 = len2(x1, y1)
    l2 = len2(x2, y2)

    if  l1 <= l2:
        xnew = x1
        ynew = y1
        l    = l1
        flag = 1

    else:
        xnew = x2
        ynew = y2
        l    = l2
        flag = -1

    return((xnew, ynew, l, flag))


n   = int(input())
v   = []
ans = [1 for i in range(n)]
r   = 1000000**2

for i in range(n):
    (x, y) = (int(j) for j in input().split())
    v.append([x,y])

start = time.time()

#v.sort(key=lambda x: x[2], reverse = True)

if n>1:
    v1 = v[0]
    s1 =  [0]
    v2 = v[1]
    s2 =  [1]

    for i in range(2, n):

        v3 = v[i]
        s3 = [i]

    #    print(v1)
    #    print(v2)
    #    print(v3)
    #    print()
        (xnew, ynew, l, flag) = test(v1, v2)
        if l > r:
            (xnew, ynew, t, flag) = test(v1, v3)
            if t > r:
                (xnew, ynew, t, flag) = test(v2, v3)

                if flag == -1:
                    for j in s3:
                        ans[j] *= -1

                v2  = [xnew, ynew]
                s2 += s3

            else:
                if flag == -1:
                    for j in s3:
                        ans[j] *= -1

                v1  = [xnew, ynew]
                s1 += s3

        else:
            if flag == -1:
                for j in s2:
                    ans[j] *= -1

            v1  = [xnew, ynew]
            s1 += s2
            v2 = v3
            s2 = s3


    (xnew, ynew, t, flag) = test(v1, v2)

    if flag == -1:
        for j in s2:
            ans[j] *= -1

for i in range(n):
    print(ans[i], end = " ")
print()
finish = time.time()
#print(finish - start)
