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 
#include <bits/stdc++.h>
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
using namespace std;
class otr {
 public:
  long long len = 0;
  long long* D = nullptr;
  otr(long long alen) {
    len = alen;
    D = new long long[alen];
  }
  ~otr() { delete[] D; }
};
void Flip(long long* left, long long* right, long long flix, bool isleft) {
  while (flix != -1) {
    long long t = left[flix];
    left[flix] = right[flix];
    right[flix] = t;
    flix = isleft ? right[flix] : left[flix];
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  std::cout.tie(0);
  std::istream* istr = &cin;
  const long long mdl = 1000000007;
  long long q;
  (*istr) >> q;
  for (long long qi = 0; qi < q; qi++) {
    long long n, m, a, b, c;
    (*istr) >> n;
    otr** data = new otr*[n - 1];
    for (long long i = 0; i < n - 1; i++) {
      (*istr) >> m;
      data[i] = new otr(m);
      for (long long j = 0; j < m; j++) (*istr) >> data[i]->D[j];
    }
    for (long long i = 0; i < n - 2; i++)
      for (long long j = i + 1; j < n - 1; j++)
        if (data[i]->len > data[j]->len) {
          otr* t = data[i];
          data[i] = data[j];
          data[j] = t;
        }
    long long skippedcnt = 0;
    long long* left = new long long[n];
    long long* right = new long long[n];
    long long start = -1;
    long long end = -1;
    long long qend = -1;
    map<long long, bool> mp;
    long long ix = 0;
    bool* skip = new bool[n - 1];
    while (ix < n - 1 && data[ix]->len == 2) {
      for (long long i = 0; i < n - 1; i++) skip[i] = false;
      skippedcnt = 0;
      for (long long i = 0; i < n; i++) {
        left[i] = -1;
        right[i] = -1;
      }
      mp.clear();
      right[data[ix]->D[0] - 1] = data[ix]->D[1] - 1;
      left[data[ix]->D[1] - 1] = data[ix]->D[0] - 1;
      long long lastskippedcnt = skippedcnt;
      skip[ix] = true;
      skippedcnt++;
      mp[data[ix]->D[0] - 1] = true;
      mp[data[ix]->D[1] - 1] = true;
      start = data[ix]->D[0] - 1;
      end = data[ix]->D[1] - 1;
      qend = end;
      bool isq = true;
      bool isok = true;
      while (skippedcnt < n - 1) {
        long long iy = 0;
        lastskippedcnt = skippedcnt;
        while (iy < n - 1) {
          if (!skip[iy]) {
            bool startisthere = false;
            bool endisthere = false;
            long long newone = -1;
            long long newonecount = 0;
            for (long long i = 0; i < data[iy]->len; i++) {
              if (data[iy]->D[i] - 1 == start) startisthere = true;
              if (data[iy]->D[i] - 1 == qend) endisthere = true;
              if (mp.find(data[iy]->D[i] - 1) == mp.end()) {
                newone = data[iy]->D[i] - 1;
                newonecount++;
              }
            }
            if (newonecount > 1) {
              iy++;
              continue;
            }
            skip[iy] = true;
            skippedcnt++;
            if (startisthere && !endisthere) {
              Flip(left, right, qend, true);
              long long t = start;
              start = qend;
              qend = t;
              if (isq) end = t;
              if (left[start] >= 0) {
                left[left[start]] = qend;
                right[qend] = left[start];
                left[start] = -1;
              }
            }
            left[newone] = end;
            right[end] = newone;
            end = newone;
            mp[newone] = true;
            if (isq && data[iy]->len == 2)
              qend = end;
            else
              isq = false;
            break;
          }
          iy++;
        }
        if (lastskippedcnt == skippedcnt) {
          isok = false;
          break;
        }
      }
      if (isok) break;
      ix++;
    }
    long long v = start;
    while (v >= 0) {
      cout << v + 1 << ' ';
      v = right[v];
    }
    cout << endl;
    for (long long i = 0; i < n - 1; i++) delete data[i];
    delete[] data;
  }
}
