Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long mod2 = 998244353;
long long mod3 = 1000003;
long long mod4 = 998244853;
long long mod5 = 1000000009;
long long inf = 1LL << 62;
double pi = 3.141592653589793238462643383279L;
double eps = 1e-14;
int dh[4] = {1, -1, 0, 0};
int dw[4] = {0, 0, 1, -1};
int ddh[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int ddw[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long c = gcd(a, b);
  return a * b / c;
}
long long Pow(long long n, long long k) {
  long long ret = 1;
  long long now = n;
  while (k > 0) {
    if (k & 1) ret *= now;
    now *= now;
    k /= 2;
  }
  return ret;
}
long long beki(long long n, long long k, long long md) {
  long long ret = 1;
  long long now = n;
  now %= md;
  while (k > 0) {
    if (k % 2 == 1) {
      ret *= now;
      ret %= md;
    }
    now *= now;
    now %= md;
    k /= 2;
  }
  return ret;
}
long long gyaku(long long n, long long md) { return beki(n, md - 2, md); }
long long popcount(long long n) {
  long long ret = 0;
  long long u = n;
  while (u > 0) {
    ret += u % 2;
    u /= 2;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  for (long long _ = 0; _ < t; _++) {
    int n;
    cin >> n;
    int cnt[n + 1];
    fill(cnt, cnt + n + 1, 0);
    set<int> st[n + 1];
    vector<int> v[n];
    for (long long i = 0; i < n - 1; i++) {
      int k;
      cin >> k;
      for (long long j = 0; j < k; j++) {
        int a;
        cin >> a;
        st[a].insert(i);
        v[i].push_back(a);
        cnt[a]++;
      }
    }
    vector<int> vvv;
    for (long long i = 1; i < n + 1; i++) {
      if (cnt[i] == 1) {
        vvv.push_back(i);
      }
    }
    vector<int> ans;
    for (long long u = 0; u < vvv.size(); u++) {
      ans.clear();
      int ok[n + 1];
      fill(ok, ok + n + 1, 0);
      bool used[n + 1];
      fill(used, used + n + 1, false);
      int rrr = 0;
      int cnt2[n + 1];
      for (long long i = 1; i < n + 1; i++) cnt2[i] = cnt[i];
      set<int> st2[n + 1];
      for (long long i = 1; i < n + 1; i++) st2[i] = st[i];
      for (long long i = 0; i < n - 1; i++) {
        int maa = -1;
        int numm = -1;
        int maaa = 0;
        for (long long j = 1; j < n + 1; j++) {
          if (used[j] == false) maaa = max(maaa, ok[j]);
          if (cnt2[j] == 1 && used[j] == false) {
            if (ok[j] > maa) {
              maa = ok[j];
              numm = j;
            }
          }
        }
        if (i == 0) numm = vvv[u];
        if (ok[numm] < maaa) {
          rrr = 1;
          break;
        }
        ans.push_back(numm);
        int pla = *st2[numm].begin();
        used[numm] = true;
        for (long long k = 0; k < v[pla].size(); k++) {
          int ne = v[pla][k];
          cnt2[ne]--;
          st2[ne].erase(pla);
          ok[ne]++;
        }
      }
      if (rrr) continue;
      for (long long i = 1; i < n + 1; i++) {
        if (used[i] == false) ans.push_back(i);
      }
      break;
    }
    reverse(ans.begin(), ans.end());
    for (long long i = 0; i < n; i++) cout << ans[i] << " ";
    cout << endl;
  }
}