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.io.*;
    import java.util.*;
    import java.math.*;
     
    public class Solution {
     
        InputStream is;
        PrintWriter o;
        /////////////////// CODED++ BY++ ++ ++ ++ BHAVYA++ ARORA++ ++ ++ ++ FROM++ JAYPEE++ INSTITUTE++ OF++ INFORMATION++ TECHNOLOGY++ ////////////////
     
    ///////////////////////// Make it work, make it right, make it fast. Make it work, make it right, make it fast. Make it work, make it right, make it fast. Make it work, make it right, make it fast. /////////////////
     
    
        
        void solve() {
        	int t=ni();
        	while(t-->0)
        	{
        		int n=ni();
        		int arr[]=nia(n);
        		
        		String ans[]=new String[n+1];
        		char[] res="abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz".toCharArray();
        		for(int i=0;i<n;i++)
        		{
        			
        			ans[i]=new String(res);
        			
        			int nu=arr[i];
        			char chan=(char)(97+((res[nu]-'a')+1)%26);
        			res[nu]=chan;

        			ans[i+1]=new String(res);

        		}
        		for(int i=0;i<=n;i++)
        		{
        			o.println(ans[i]);
        		}

        	}
            
            
     
            
     
     
        }
     
     
     
     
     
     
       
     
     
        
     
     
     
     
     
     
     
     
     
     
     
     
        
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
        //---------- I/O Template ----------
        
        public static void main(String[] args) { new Solution().run(); }
        void run() { 
            is = System.in; 
            o = new PrintWriter(System.out);
            solve();
            o.flush();
        }
        
        byte input[] = new byte[1024];
        int len = 0, ptr = 0;
        
        int readByte() { 
            if(ptr >= len) { ptr = 0; 
                try { len = is.read(input); } 
                catch(IOException e) { throw new InputMismatchException(); } 
                if(len <= 0) { return -1; } 
            } return input[ptr++];
        }
        boolean isSpaceChar(int c) { return !( c >= 33 && c <= 126 ); }
        int skip() { 
            int b = readByte(); 
            while(b != -1 && isSpaceChar(b)) { b = readByte(); } 
            return b;
        }
        
        char nc() { return (char)skip(); }
        String ns() { 
            int b = skip(); 
            StringBuilder sb = new StringBuilder(); 
            while(!isSpaceChar(b)) { sb.appendCodePoint(b); b = readByte(); } 
            return sb.toString();
        }
        String nLine() { 
            int b = skip(); 
            StringBuilder sb = new StringBuilder(); 
            while( !(isSpaceChar(b) && b != ' ') ) { sb.appendCodePoint(b); b = readByte(); } 
            return sb.toString();
        }
        int ni() { 
            int n = 0, b = readByte(); 
            boolean minus = false; 
            while(b != -1 && !( (b >= '0' && b <= '9') || b == '-')) { b = readByte(); } 
            if(b == '-') { minus = true; b = readByte(); } 
            if(b == -1) { return -1; }  //no input 
            while(b >= '0' && b <= '9') { n = n * 10 + (b - '0'); b = readByte(); } 
            return minus ? -n : n;
        }
        long nl() { 
            long n = 0L;    int b = readByte(); 
            boolean minus = false; 
            while(b != -1 && !( (b >= '0' && b <= '9') || b == '-')) { b = readByte(); } 
            if(b == '-') { minus = true; b = readByte(); } 
            while(b >= '0' && b <= '9') { n = n * 10 + (b - '0'); b = readByte(); } 
            return minus ? -n : n;
        }
     
        double nd() { return Double.parseDouble(ns()); }
        float nf() { return Float.parseFloat(ns()); }
        int[] nia(int n) { 
            int a[] = new int[n]; 
            for(int i = 0; i < n; i++) { a[i] = ni(); } 
            return a;
        }
        long[] nla(int n) { 
            long a[] = new long[n]; 
            for(int i = 0; i < n; i++) { a[i] = nl(); } 
            return a;
        }
        int [][] nim(int n)
        {
            int mat[][]=new int[n][n];
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<n;j++)
                {
                    mat[i][j]=ni();
                }
            }
            return mat;
        }
        long [][] nlm(int n)
        {
            long mat[][]=new long[n][n];
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<n;j++)
                {
                    mat[i][j]=nl();
                }
            }
            return mat;
        }
        
     
     
     
     
        
        char[] ns(int n) { 
            char c[] = new char[n]; 
            int i, b = skip(); 
            for(i = 0; i < n; i++) { 
                if(isSpaceChar(b)) { break; } 
                c[i] = (char)b; b = readByte(); 
            } return i == n ? c : Arrays.copyOf(c,i);
        }
        void piarr(int arr[])
        {
            for(int i=0;i<arr.length;i++)
            {
                o.print(arr[i]+" ");
     
            }
            o.println();
        }
        void plarr(long arr[])
        {
            for(int i=0;i<arr.length;i++)
            {
                o.print(arr[i]+" ");
     
            }
            o.println();
        }
        
        void pimat(int mat[][])
        {
            for(int i=0;i<mat.length;i++)
            {
                for(int j=0;j<mat[0].length;j++)
                {
                    o.print(mat[i][j]+" ");
                }
                o.println();
            }
        }
        void plmat(long mat[][])
        {
            for(int i=0;i<mat.length;i++)
            {
                for(int j=0;j<mat[0].length;j++)
                {
                    o.print(mat[i][j]+" ");
                }
                o.println();
            }
     
        }
       static class DJset
{
    int n;
    ty1 arr[];
    class ty1
    {
        int rank;
        int par;
        ty1(int p,int r)
        {
            par=p;
            rank=r;
        }
    }
    DJset(int nu)
    {
        n=nu;
        arr=new ty1[n];
        for(int i=0;i<n;i++)
        {
            arr[i]=new ty1(i,1);
        }

    }
    int find(int x)
    {
        if(arr[x].par!=x)
        {
            arr[x].par=find(arr[x].par);
        }
        return arr[x].par;
    }
    void union(int x,int y)
    {
        x=find(x);
        y=find(y);
        if(x!=y)
        {
            if(arr[x].rank<arr[y].rank)
            {
                arr[x].par=y;
            }
            else if(arr[y].rank<arr[x].rank)
            {
                arr[y].par=x;
            }
            else
            {
                arr[x].rank++;
                arr[y].par=x;

            }
        }
    }
    public String toString()
    {
        String res="";
        for(int i=0;i<n;i++)
        {
            res+=(arr[i].par+" ");
        }
        
        return res;
    }
    public int countsets()
    {
        int res=0;
        for(int i=0;i<n;i++)
        {
            if(arr[i].par==i)
            {
                res++;
            }
        }
        return res;
    }

    

    }
    
     
     
     
     
        //////////////////////////////////// template finished //////////////////////////////////////
     
     
     
    }