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;
const double PI = acos(-1);
const double eps = 1e-12;
const int inf = 2000000000;
const long long int infLL = (long long int)1e18;
long long int MOD = 1000000007;
int MOD1 = 1000000007;
int MOD2 = 1000000009;
inline bool checkBit(long long int n, long long int i) {
  return n & (1LL << i);
}
inline long long int setBit(long long int n, long long int i) {
  return n | (1LL << i);
  ;
}
inline long long int resetBit(long long int n, long long int i) {
  return n & (~(1LL << i));
}
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long int year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline void normal(long long int &a) {
  a %= MOD;
  (a < 0) && (a += MOD);
}
inline long long int modMul(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a * b) % MOD;
}
inline long long int modAdd(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a + b) % MOD;
}
inline long long int modSub(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long int modPow(long long int b, long long int p) {
  long long int r = 1LL;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1LL;
  }
  return r;
}
inline long long int modDiv(long long int a, long long int b) {
  return modMul(a, modPow(b, MOD - 2));
}
bool comp(const pair<long long int, pair<long long int, long long int> > &p1,
          const pair<long long int, pair<long long int, long long int> > &p2) {
  return p1.first > p2.first;
}
bool comp1(const pair<long long int, long long int> &p1,
           const pair<long long int, long long int> &p2) {
  if (p1.first == p2.first) {
    return p1.second > p2.second;
  }
  return p1.first < p2.first;
}
long long int converter(string a) {
  long long int i, mul = 1LL, r, t, ans = 0LL;
  if (a.length() == 0) return 0;
  for (i = a.length() - 1; i >= 0; i--) {
    t = a[i] - '0';
    r = t % 10;
    ans += (mul * r);
    mul = mul * 10;
  }
  return ans;
}
int msb(unsigned x) {
  union {
    double a;
    int b[2];
  };
  a = x;
  return (b[1] >> 20) - 1023;
}
const int MAX = 205;
set<int> s[MAX];
set<int> t[MAX];
int n;
vector<int> store;
int pos[MAX];
int process(int st) {
  memset(pos, 0, sizeof(pos));
  for (int i = 1; i < n; ++i) t[i] = s[i];
  store.clear();
  for (int i = 1; i < n; ++i) {
    if (t[i].find(st) != t[i].end()) t[i].erase(st);
  }
  store.push_back(st);
  pos[st] = 1;
  while (1) {
    int idx, cnt = 0;
    for (int i = 1; i < n; ++i) {
      if (t[i].size() == 0 || t[i].size() > 1)
        continue;
      else {
        cnt++;
        idx = i;
      }
    }
    if (cnt == 0)
      break;
    else if (cnt > 1)
      return 0;
    int cur = *t[idx].begin();
    store.push_back(*t[idx].begin());
    pos[cur] = store.size();
    for (int i = 1; i < n; ++i)
      if (t[i].find(cur) != t[i].end()) {
        t[i].erase(cur);
      }
  }
  for (int i = 1; i < n; ++i) {
    int mini = inf, maxi = 0;
    for (auto x : s[i]) {
      mini = min(pos[x], mini);
      maxi = max(pos[x], maxi);
    }
    if (maxi - mini + 1 == s[i].size())
      ;
    else
      return 0;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cout.unsetf(ios::floatfield);
  cout.precision(20);
  cout.setf(ios::fixed, ios::floatfield);
  ;
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; ++i) s[i].clear();
    for (int i = 1; i < n; ++i) {
      int k, x;
      cin >> k;
      for (int j = 1; j <= k; ++j) {
        cin >> x;
        s[i].insert(x);
      }
    }
    for (int i = 1; i <= n; ++i) {
      if (process(i)) {
        break;
      }
    }
    for (auto x : store) cout << x << " ";
    cout << '\n';
  }
  return 0;
}