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 PYTHON):

Solution:
#!/usr/bin/env python
"""
This file is part of https://github.com/Cheran-Senthil/PyRival.

Copyright 2018 Cheran Senthilkumar all rights reserved,
Cheran Senthilkumar <hello@cheran.io>
Permission to use, modify, and distribute this software is given under the
terms of the MIT License.

"""
from __future__ import division, print_function

import cmath
import itertools
import math
import operator as op
# import random
import sys
from atexit import register
from bisect import bisect_left, bisect_right
# from collections import Counter, MutableSequence, defaultdict, deque
# from copy import deepcopy
# from decimal import Decimal
# from difflib import SequenceMatcher
# from fractions import Fraction
# from heapq import heappop, heappush
import io


if sys.version_info[0] < 3:
    # from cPickle import dumps
    # from Queue import PriorityQueue, Queue

    class dict(dict):
        """dict() -> new empty dictionary"""
        def items(self):
            """D.items() -> a set-like object providing a view on D's items"""
            return dict.iteritems(self)

        def keys(self):
            """D.keys() -> a set-like object providing a view on D's keys"""
            return dict.iterkeys(self)

        def values(self):
            """D.values() -> an object providing a view on D's values"""
            return dict.itervalues(self)

    input = iter(io.FileIO(0).read().splitlines()).next
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip

    sys.stdout = io.BytesIO()
    register(lambda: io.FileIO(1, 'w').write(sys.stdout.getvalue()))

else:
    # from functools import reduce
    # from pickle import dumps
    # from queue import PriorityQueue, Queue

    input = iter(io.FileIO(0).read().splitlines()).__next__

    sys.stdout = io.StringIO()
    register(lambda: io.FileIO(1, 'w').write(sys.stdout.getvalue().encode()))


def gcd(x, y):
    """greatest common divisor of x and y"""
    while y:
        x, y = y, x % y
    return x


def main():
    for _ in range(int(input())):
        n = int(input())
        q = sorted([list(map(int, input().split())) + [i] for i in range(n)])

        t = [2] * n

        le = q[0][1]

        for s, e, i in q:
            if s > le: break

            t[i] = 1
            le = max(le, e)

        if t[q[-1][2]] == 1:
            print(-1)
            continue

        print(*t)


if __name__ == '__main__':
    main()