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 JAVA):

Solution:
import java.util.*;
import java.io.*;
public class Main implements Runnable{
    FastScanner sc;
    PrintWriter pw;
    final class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            try {
                br = new BufferedReader(new InputStreamReader(System.in));
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
 
        public long nlo() {
            return Long.parseLong(next());
        }
 
        public String next() {
            if (st.hasMoreTokens()) return st.nextToken();
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
            return st.nextToken();
        }
 
        public int ni() {
            return Integer.parseInt(next());
        }
 
        public String nli() {
            String line = "";
            if (st.hasMoreTokens()) line = st.nextToken();
            else try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            while (st.hasMoreTokens()) line += " " + st.nextToken();
            return line;
        }
 
        public double nd() {
            return Double.parseDouble(next());
        }
    }
    public static void main(String[] args) throws Exception
    {
        new Thread(null,new Main(),"codeforces",1<<28).start();
    }
    public void run()
    {
        sc=new FastScanner();
        pw=new PrintWriter(System.out);
        solve();
        pw.flush();
        pw.close();
    }
    public long gcd(long a,long b)
    {
        return b==0L?a:gcd(b,a%b);
    }
    public long ppow(long a,long b,long mod)
    {
        if(b==0L)
        return 1L;
        long tmp=1;
        while(b>1L)
        {
            if((b&1L)==1)
            tmp*=a;
            a*=a;
            a%=mod;
            tmp%=mod;
            b>>=1;
        }
        return (tmp*a)%mod;
    }
    public  int gcd(int x,int y)
    {
        return y==0?x:gcd(y,x%y);
    }
    public class Pair implements Comparable<Pair>{
        int x;
        int val;
        Pair(int a,int b)
        {
            x=a;
            val=b;
        }
        public int compareTo(Pair b)
        {
            return this.val-b.val;
        }
        public Pair clone(Pair b)
        {
            return new Pair(b.x,b.val);
        }
    }
    //////////////////////////////////
    /////////////  LOGIC  ///////////
    ////////////////////////////////

    public void solve() {
        int t=sc.ni();
        while(t-->0)
        {
           int n=sc.ni();
           int[] arr=new int[n];
           for(int i=0;i<n;i++)
           arr[i]=sc.ni();
           int p=0,l=200;
           StringBuilder str[] =new StringBuilder[n+1];
           for(int i=0;i<=n;i++)
           str[i]=new StringBuilder();
           StringBuilder tmp=new StringBuilder();
           for(int i=0;i<50;i++)
           tmp.append("a");
           str[n].append(tmp);
           for(int i=n-1;i>=0;i--)
           {
               l=str[i+1].length();
            
                if(arr[i]==0)
                {
                    char ch=(char)((p+1)%26+'a');
                    String x=Character.toString(ch);
                    str[i].append(x);
                }
                else if(arr[i]<=l)
                    str[i].append(str[i+1].substring(0,arr[i]));
                else
                {
                    str[i].append(str[i+1].substring(0,l));
                    char ch=(char)((p+1)%26+'a');
                    String x=Character.toString(ch);
                    if(i+2<=n&&str[i+2].length()>str[i].length())
                    {
                        ch=str[i+2].charAt(str[i].length());
                        ch=(char)(((int)(ch-'a')+1)%26+'a');
                        x=Character.toString(ch);
                    }
                    for(int j=l+1;j<=arr[i];j++)
                    {str[i].append(x);str[i+1].append(x);}
                }
                char ch=str[i].charAt(str[i].length()-1);
                p=ch-'a';
           }
           for(int i=0;i<=n;i++)
           pw.println(str[i]);
        }
    }
}