Problem: The length of the longest common prefix of two strings s = s_1 s_2 … s_n and t = t_1 t_2 … t_m is defined as the maximum integer k (0 ≤ k ≤ min(n,m)) such that s_1 s_2 … s_k equals t_1 t_2 … t_k.

Koa the Koala initially has n+1 strings s_1, s_2, ..., s_{n+1}.

For each i (1 ≤ i ≤ n) she calculated a_i — the length of the longest common prefix of s_i and s_{i+1}.

Several days later Koa found these numbers, but she couldn't remember the strings.

So Koa would like to find some strings s_1, s_2, ..., s_{n+1} which would have generated numbers a_1, a_2, ..., a_n. Can you help her?

If there are many answers print any. We can show that answer always exists for the given constraints. 

Input

Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 100) — the number of test cases. Description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 100) — the number of elements in the list a.

The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 50) — the elements of a.

It is guaranteed that the sum of n over all test cases does not exceed 100.

Output

For each test case:

Output n+1 lines. In the i-th line print string s_i (1 ≤ |s_i| ≤ 200), consisting of lowercase Latin letters. Length of the longest common prefix of strings s_i and s_{i+1} has to be equal to a_i.

If there are many answers print any. We can show that answer always exists for the given constraints.

Example

Input


4
4
1 2 4 2
2
5 3
3
1 3 1
3
0 0 0


Output


aeren
ari
arousal
around
ari
monogon
monogamy
monthly
kevinvu
kuroni
kurioni
korone
anton
loves
adhoc
problems

Note

In the 1-st test case one of the possible answers is s = [aeren, ari, arousal, around, ari].

Lengths of longest common prefixes are:

  * Between \color{red}{a}eren and \color{red}{a}ri → 1 
  * Between \color{red}{ar}i and \color{red}{ar}ousal → 2 
  * Between \color{red}{arou}sal and \color{red}{arou}nd → 4 
  * Between \color{red}{ar}ound and \color{red}{ar}i → 2 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, long long p) { return power(n, p - 2, p); }
long long nCrModPFermat(long long n, long long r, long long p) {
  if (r == 0) return 1;
  long long fac[n + 1];
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s[26];
  s[0] = "a";
  for (long long i = 1; i < 26; i++) {
    s[i] = s[i - 1] + (char)(i + 'a');
  }
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    string ans[n + 1];
    long long k = 0;
    for (long long i = 0; i < n; i++) {
      if (a[i] == 0) {
        if (ans[i].size() == 0) ans[i] += ('a' + k++);
        k %= 26;
        ans[i + 1] += ('a' + k++);
      } else if (a[i] >= ans[i].size()) {
        a[i] -= ans[i].size();
        while (a[i]--) {
          ans[i] += ('a' + k);
        }
        k++;
        k = k % 26;
        ans[i + 1] = ans[i];
      } else {
        for (long long j = 0; j < a[i]; j++) {
          ans[i + 1] += ans[i][j];
        }
      }
      k %= 26;
    }
    for (long long i = 0; i <= n; i++) {
      cout << ans[i] << endl;
    }
  }
}