Problem: All techniques in the ninja world consist of hand seals. At the moment Naruto is learning a new technique, which consists of n⋅ m different seals, denoted by distinct numbers. All of them were written in an n× m table.

The table is lost now. Naruto managed to remember elements of each row from left to right, and elements of each column from top to bottom, but he doesn't remember the order of rows and columns. Please restore the table consistent with this data so that Naruto will be able to learn the new technique.

Input

The first line of the input contains the only integer t (1≤ t≤ 100 000) denoting the number of test cases. Their descriptions follow.

The first line of each test case description consists of two space-separated integers n and m (1 ≤ n, m ≤ 500) standing for the number of rows and columns in the table, respectively. All hand seals are encoded by the positive integers from 1 to n⋅ m.

The following n lines contain m space separated integers each, denoting elements of an arbitrary row in the table left to right.

The following m lines contain n space separated integers each, denoting elements of an arbitrary column in the table top to bottom.

Sum of nm over all test cases does not exceed 250 000. It is guaranteed that each row occurs in the input exactly once, as well as each column. It is also guaranteed that each number from 1 to nm occurs exactly once in all rows, as well as in all columns. Finally, it is guaranteed that a table consistent with the input exists.

Output

For each test case, output n lines with m space-separated integers each, denoting the restored table. One can show that the answer is always unique.

Example

Input


2
2 3
6 5 4
1 2 3
1 6
2 5
3 4
3 1
2
3
1
3 1 2


Output


1 2 3 
6 5 4 
3 
1 
2 

Note

Consider the first test case. The matrix is 2 × 3. You are given the rows and columns in arbitrary order.

One of the rows is [6, 5, 4]. One of the rows is [1, 2, 3].

One of the columns is [1, 6]. One of the columns is [2, 5]. One of the columns is [3, 4].

You are to reconstruct the matrix. The answer is given in the output.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long binpow(long long a, long long n, long long mod) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    n >>= 1;
  }
  return ans;
}
long long binpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    a *= a;
    n >>= 1;
  }
  return ans;
}
vector<long long> pr;
void PrDoX(long long x) {
  if (pr.size() > 0) return;
  vector<long long> p(x + 1);
  for (long long i = 2; i <= x; i++) {
    if (p[i] == 0) {
      pr.push_back(i);
      for (long long j = i + i; j <= x; j += i) p[j] = 1;
    }
  }
}
vector<long long> fact(long long x) {
  PrDoX(sqrt(1e9));
  vector<long long> ans;
  long long i = 0, c = sqrt(x), n = pr.size();
  while (x > 1 and pr[i] <= c and i < n) {
    if (x % pr[i] == 0) {
      x /= pr[i];
      ans.push_back(pr[i]);
      c = sqrt(x);
    } else
      i++;
  }
  if (x != 1) ans.push_back(x);
  return ans;
}
long long n, t, a[501][501], b[501][501], k, l, o, m;
int main() {
  cin >> t;
  while (t--) {
    cin >> n >> m;
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < m; j++) cin >> a[i][j];
    }
    for (long long i = 0; i < m; i++) {
      for (long long j = 0; j < n; j++) cin >> b[i][j];
    }
    long long l, x;
    for (long long i = 0; i < n; i++) {
      bool fl = 0;
      for (long long j = 0; j < m; j++) {
        if (a[i][j] == b[0][0]) {
          l = i;
          fl = 1;
          break;
        }
      }
      if (fl == 1) break;
    }
    vector<pair<long long, long long>> v(m);
    for (long long i = 0; i < m; i++) {
      for (long long j = 0; j < m; j++) {
        if (a[l][i] == b[j][0]) {
          v[i] = {i, j};
          break;
        }
      }
    }
    sort(v.begin(), v.end());
    vector<vector<long long>> v1(m, vector<long long>(n));
    for (long long i = 0; i < m; i++) {
      for (long long j = 0; j < n; j++) {
        v1[i][j] = b[v[i].second][j];
      }
    }
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < m; j++) {
        cout << v1[j][i] << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}