We guessed a permutation $p$ consisting of $n$ integers. The permutation of length $n$ is the array of length $n$ where each element from $1$ to $n$ appears exactly once. This permutation is a secret for you.

For each position $r$ from $2$ to $n$ we chose some other index $l$ ($l < r$) and gave you the segment $p_l, p_{l + 1}, \dots, p_r$ in sorted order (i.e. we rearranged the elements of this segment in a way that the elements of this segment are sorted). Thus, you are given exactly $n-1$ segments of the initial permutation but elements inside each segment are sorted. The segments are given to you in random order.

For example, if the secret permutation is $p=[3, 1, 4, 6, 2, 5]$ then the possible given set of segments can be:  $[2, 5, 6]$  $[4, 6]$  $[1, 3, 4]$  $[1, 3]$  $[1, 2, 4, 6]$ 

Your task is to find any suitable permutation (i.e. any permutation corresponding to the given input data). It is guaranteed that the input data corresponds to some permutation (i.e. such permutation exists).

You have to answer $t$ independent test cases.


-----Input-----

The first line of the input contains one integer $t$ ($1 \le t \le 100$) — the number of test cases. Then $t$ test cases follow.

The first line of the test case contains one integer $n$ ($2 \le n \le 200$) — the length of the permutation.

The next $n-1$ lines describe given segments.

The $i$-th line contains the description of the $i$-th segment. The line starts with the integer $k_i$ ($2 \le k_i \le n$) — the length of the $i$-th segment. Then $k_i$ integers follow. All integers in a line are distinct, sorted in ascending order, between $1$ and $n$, inclusive.

It is guaranteed that the required $p$ exists for each test case.

It is also guaranteed that the sum of $n$ over all test cases does not exceed $200$ ($\sum n \le 200$).


-----Output-----

For each test case, print the answer: $n$ integers $p_1, p_2, \dots, p_n$ ($1 \le p_i \le n$, all $p_i$ should be distinct) — any suitable permutation (i.e. any permutation corresponding to the test case input).


-----Example-----
Input
5
6
3 2 5 6
2 4 6
3 1 3 4
2 1 3
4 1 2 4 6
5
2 2 3
2 1 2
2 1 4
2 4 5
7
3 1 2 6
4 1 3 5 6
2 1 2
3 4 5 7
6 1 2 3 4 5 6
3 1 3 6
2
2 1 2
5
2 2 5
3 2 3 5
4 2 3 4 5
5 1 2 3 4 5

Output
3 1 4 6 2 5 
3 2 1 4 5 
2 1 6 3 5 4 7 
1 2 
2 5 3 4 1

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = "from collections import Counter\nfrom itertools import chain\n\n\ndef dfs(n, r, hint_sets, count, removed, result):\n    # print(n, r, hint_sets, count, removed, result)\n    if len(result) == n - 1:\n        last = (set(range(1, n + 1)) - set(result)).pop()\n        result.append(last)\n        return True\n\n    i, including_r = 0, None\n    for i, including_r in enumerate(hint_sets):\n        if i in removed:\n            continue\n        if r in including_r:\n            break\n    removed.add(i)\n\n    next_r = []\n    for q in including_r:\n        count[q] -= 1\n        if count[q] == 1:\n            next_r.append(q)\n    if not next_r:\n        return False\n\n    # print(r, next_r, result)\n\n    if len(next_r) == 2:\n        nr = -1\n        can1, can2 = next_r\n        for h in hint_sets:\n            if can1 in h and can2 not in h and not h.isdisjoint(result):\n                nr = can1\n                break\n            if can1 not in h and can2 in h and not h.isdisjoint(result):\n                nr = can2\n                break\n        if nr == -1:\n            nr = can1\n    else:\n        nr = next_r[0]\n\n    result.append(nr)\n    res = dfs(n, nr, hint_sets, count, removed, result)\n    if res:\n        return True\n    result.pop()\n\n    for q in including_r:\n        count[q] += 1\n\n    return False\n\n\nt = int(input())\nbuf = []\nfor l in range(t):\n    n = int(input())\n    hints = []\n    for _ in range(n - 1):\n        k, *ppp = list(map(int, input().split()))\n        hints.append(ppp)\n\n    count = Counter(chain.from_iterable(hints))\n    most_common = count.most_common()\n\n    hint_sets = list(map(set, hints))\n\n    r = most_common[-1][0]\n    result = [r]\n    if dfs(n, r, hint_sets, dict(count), set(), result):\n        buf.append(' '.join(map(str, result[::-1])))\n        continue\n    r = most_common[-2][0]\n    result = [r]\n    assert dfs(n, r, hint_sets, dict(count), set(), result)\n    buf.append(' '.join(map(str, result[::-1])))\n\nprint('\\n'.join(map(str, buf)))\n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 20619