Problem: There are n segments [l_i, r_i] for 1 ≤ i ≤ n. You should divide all segments into two non-empty groups in such way that there is no pair of segments from different groups which have at least one common point, or say that it's impossible to do it. Each segment should belong to exactly one group.

To optimize testing process you will be given multitest.

Input

The first line contains one integer T (1 ≤ T ≤ 50000) — the number of queries. Each query contains description of the set of segments. Queries are independent.

First line of each query contains single integer n (2 ≤ n ≤ 10^5) — number of segments. It is guaranteed that ∑{n} over all queries does not exceed 10^5.

The next n lines contains two integers l_i, r_i per line (1 ≤ l_i ≤ r_i ≤ 2 ⋅ 10^5) — the i-th segment.

Output

For each query print n integers t_1, t_2, ..., t_n (t_i ∈ \{1, 2\}) — for each segment (in the same order as in the input) t_i equals 1 if the i-th segment will belongs to the first group and 2 otherwise.

If there are multiple answers, you can print any of them. If there is no answer, print -1.

Example

Input


3
2
5 5
2 3
3
3 5
2 3
2 3
3
3 3
4 4
5 5


Output


2 1 
-1
1 1 2 

Note

In the first query the first and the second segments should be in different groups, but exact numbers don't matter.

In the second query the third segment intersects with the first and the second segments, so they should be in the same group, but then the other group becomes empty, so answer is -1.

In the third query we can distribute segments in any way that makes groups non-empty, so any answer of 6 possible is correct.

Solution (in PYTHON3):

Solution:
#CodeForces Educational Round 58
#C Division and Union - the other's code

query = int(input())
for _ in range(query):
    n = int(input())
    array = []
    for i in range(n):
        a, b = map(int, input().split())
        array.append([a, b, i])
    array.sort() #sort the list of lists in ascending order
    
    max1 = array[0][1] #a boundary
    flag = False #Ture means there is something in the second group
    answer = [0] * n
    answer[array[0][2]] = 1

    for i in range(1, n):
        #start from the smallest, check if it overlaps with the boundary
        #place it in group one if it does, and shift the boundary
        if (not flag) and (array[i][0] <= max1):
            max1 = max(max1, array[i][1])
            answer[array[i][2]] = 1

        #otherwise, place it and all the rest of the segments in group two.
        #There is no need to check for overlaps.
        else:
            flag = True
            answer[array[i][2]] = 2
    if not flag:
        print(-1)
    else:
        print(*answer)