// Model solution for problem K
// Author: Adam Polak

#include <iostream>
#include <string>
using namespace std;

const int N = 26;

bool graph[N][N];
int color[N], dp[N], parent[N];

void dfs(int u, int f) {
  if (color[u] == 2) return;
  parent[u] = f;
  if (color[u] == 1) throw u;
  color[u] = 1;
  parent[u] = f;
  for (int v = 0; v < N; ++v)
    if (graph[u][v]) {
      dfs(v, u);
      dp[u] = max(dp[u], 1 + dp[v]);
    }
  color[u] = 2;
}

int main() {
  ios_base::sync_with_stdio(0);
  int Z;
  cin >> Z;
  while (Z--) {
    for (int i = 0; i < N; ++i)
      for (int j = 0; j < N; ++j)
        graph[i][j] = true;
    int n;
    cin >> n;
    while (n--) {
      string w;
      cin >> w;
      graph[w[0]-'a'][w[1]-'a'] = false;
    }
    for (int i = 0; i < N; ++i) {
      color[i] = 0;
      dp[i] = 0;
    }
    int on_cycle = -1;
    try {
      for (int i = 0; i < N; ++i)
        dfs(i, -1);
    } catch (int ex_on_cycle) {
      on_cycle = ex_on_cycle;
    }
    int path[100], path_len = 0;
    if (on_cycle != -1) {
      path_len = 39;  // this makes 20x20 square
      for (int i = path_len - 1; i >= 0; --i) {
        path[i] = on_cycle;
        on_cycle = parent[on_cycle];
      }
    } else {
      int u = 0;
      for (int i = 0; i < N; ++i)
        if (dp[i] > dp[u]) u = i;
      while (dp[u] > 0) {
        path[path_len++] = u;
        for (int v = 0; v < N; ++v)
          if (graph[u][v] && dp[v] == dp[u] - 1) {
            u = v;
            break;
          }
      }
      path[path_len++] = u;
    }
    path_len = (path_len + 1) / 2;
    for (int i = 0; i < path_len; ++i) {
      for (int j = 0; j < path_len; ++j)
        cout << static_cast<char>('a' + path[i + j]);
      cout << endl;
    }
  }
  return 0;
}