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}, ..., 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 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (2 ≤ n ≤ 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 ≤ k_i ≤ 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 (∑ n ≤ 200).

Output

For each test case, print the answer: n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ 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 
import io
import os
from collections import defaultdict
from copy import deepcopy




def solve(N, segments):

    def greedy(ans, memToSeg, first):
        # If a number is only in a single segment, it's either the first or last element
        # We need to be able to pick the last element for us to inductively remove a segment and look for the next element that is only contained in one segment
        # If there's more than one candidate, try both as the potential first
        # Initially our first is unconstrained
        while len(memToSeg) > 2:
            candidates = [k for k, v in memToSeg.items() if len(v) == 1 and k != first]
            if len(candidates) != 1:
                if len(candidates) == 0:
                    # If we found zero candidates, it means our choice of first is invalid
                    assert first is not None
                    return None
                # Otherwise we found our candidates for first and last
                assert len(candidates) == 2 and first is None
                for c in candidates:
                    # Branch the greedy to try both cases. This will only happen once
                    temp = greedy(deepcopy(ans), deepcopy(memToSeg), first=c)
                    if temp is None:
                        continue
                    # Double check that our answer works
                    numToIndex = {x: i for i, x in enumerate(temp)}
                    bad = False
                    for segment in segments:
                        mn = min(numToIndex[x] for x in segment)
                        mx = max(numToIndex[x] for x in segment)
                        if mx - mn + 1 != len(segment):
                            bad = True
                            break
                    if not bad:
                        return temp
                assert False
            else:
                # If we only had one candidate we can continue inductively removing it as the last element
                assert len(candidates) == 1
                (k,) = candidates
                v = memToSeg[candidates[0]]
                assert len(v) == 1
                (index,) = v
                ans.append(k)
                for x in segments[index]:
                    memToSeg[x].remove(index)
                    if len(memToSeg[x]) == 0:
                        assert x == k
                        del memToSeg[x]    
        # In the very end we have a single segment left of length 2 and both remaining numbers point to it
        firstTwo = []
        assert len(memToSeg) == 2
        for k, v in memToSeg.items():
            assert len(v) == 1
            firstTwo.append(k)
        
        if first is not None:
            assert first in firstTwo
            if firstTwo[0] == first:
                ans.append(firstTwo[1])
                ans.append(firstTwo[0])
            else:
                ans.append(firstTwo[0])
                ans.append(firstTwo[1])
        else:
            for segment in segments:
                if firstTwo[0] in segment and firstTwo[1] not in segment:
                    ans.append(firstTwo[0])
                    ans.append(firstTwo[1])
                    break
                if firstTwo[1] in segment and firstTwo[0] not in segment:
                    ans.append(firstTwo[1])
                    ans.append(firstTwo[0])
                    break
            else:
                ans.extend(firstTwo)
        return ans

    memToSeg = defaultdict(set)
    for i, segment in enumerate(segments):
        for x in segment:
            memToSeg[x].add(i)
    return " ".join(map(str, reversed(greedy([], memToSeg, None))))



if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    T = int(input())
    for t in range(T):
        (N,) = [int(x) for x in input().split()]
        segments = []
        for i in range(N - 1):
            segment = [int(x) for x in input().split()]
            assert segment[0] == len(segment) - 1
            segments.append(segment[1:])

        ans = solve(N, segments)
        print(ans)
