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.*;
import java.math.*;
public class Main{
	static long mod = (long)(1e9+7);
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	public static void main(String sp[])throws IOException{
		Scanner sc = new Scanner(System.in);
		//FastReader sc = new FastReader();	
		int t = sc.nextInt();
		while(t-->0){
			int n = sc.nextInt();
			char arr[] = new char[100];
			for(int j=0;j<100;j++)
				arr[j]='a';
			
			for(int i=0;i<n;i++){
				if(i==0){
					String y = new String(arr);
					System.out.println(y);
				}
				int x = sc.nextInt();
				if(arr[x]=='z')
					arr[x]='a';
				else{
					arr[x] = (char)(arr[x]+1);
				}
				String u = new String(arr);
				System.out.println(u);
			}
		}
		
		
		
		
	}
	
	
	public static ArrayList<Integer> primefactor(int n){
		int sqrt = (int)Math.sqrt(n);
		ArrayList<Integer> al = new ArrayList<>();
		while(n%2==0){
			al.add(2);
			n/=2;
		}
		for(int i=3;i<=sqrt;i+=2){
			while(n%i==0){
				al.add(i);
				n/=i;
			}
		}
		if(n>2)
			al.add(n);
		return al;
	}
	
	public static long sum(long val){
		long fg =0 ;
		while(val!=0){
			fg+=val%10;
			val/=10;
		}
		return fg;
	}
	
	
	
	public static ArrayList<Integer> factor(int n){
		int sqrt = (int)Math.sqrt(n);
		ArrayList<Integer> al = new ArrayList<>();
		for(int i=1;i<=sqrt;i++){
			if(n%i!=0)
				continue;
			int first = i;
			int second = n/i;
			al.add(i);
			if(first==second){
				continue;
			}
			al.add(n/i);
		}
		return al;
	}
	
	public static long power(long x, long y) 
	{ 
		long temp; 
		if( y == 0) 
			return 1l; 
		temp = power(x, y/2); 
		if (y%2 == 0) 
			return (temp*temp)%mod; 
		else
			return ((x*temp)%mod*temp)%mod; 
	} 
	
	public static ArrayList<Integer> comp(){
		ArrayList<Integer> al = new ArrayList<>();
			int n = (int)2e5;
			boolean arr[] = new boolean[n+1];
			int sqrt = (int)Math.sqrt(n);
			for(int i=2;i<=sqrt;i++){
				if(arr[i]==false){
					for(int j=i*i;j<=n;j+=i){
						arr[j]=true;
					}
				}
			}
			for(int i=2;i<=n;i++){
				if(arr[i]==false)
					al.add(i);
			}
			
		return al;
	}
	
	public static class pair{
		long x;
		long y;
		long diff;
		pair(long x, long y, long d){
			this.x=x;
			this.y=y;
			diff = d;
		}
	}
	
	public static class comp implements Comparator<pair>{
		public int compare(pair o1, pair o2){
			return Long.valueOf(o1.diff).compareTo(Long.valueOf(o2.diff));
		}
	}
	
	
	
	static class Node{
		int node;
		int d;
		ArrayList<Integer> al = new ArrayList<>();	
	}
	

	static int gcd(int a, int b) 
    { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }
	
		
	static class FastReader 
	{ 
		final private int BUFFER_SIZE = 1 << 16; 
		private DataInputStream din; 
		private byte[] buffer; 
		private int bufferPointer, bytesRead; 
  
		public FastReader() 
		{ 
			din = new DataInputStream(System.in); 
			buffer = new byte[BUFFER_SIZE]; 
			bufferPointer = bytesRead = 0; 
		} 
  
		public FastReader(String file_name) throws IOException 
		{ 
			din = new DataInputStream(new FileInputStream(file_name)); 
			buffer = new byte[BUFFER_SIZE]; 
			bufferPointer = bytesRead = 0; 
		} 
  
		public String readLine() throws IOException 
		{ 
			byte[] buf = new byte[64]; // line length 
			int cnt = 0, c; 
			while ((c = read()) != -1) 
			{ 
				if (c == '\n') 
					break; 
				buf[cnt++] = (byte) c; 
			} 
			return new String(buf, 0, cnt); 
		} 
  
		public int nextInt() throws IOException 
		{ 
			int ret = 0; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 
			do
			{ 
				ret = ret * 10 + c - '0'; 
			}  while ((c = read()) >= '0' && c <= '9'); 
  
			if (neg) 
				return -ret; 
			return ret; 
		} 
  
		public long nextLong() throws IOException 
		{ 
			long ret = 0; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 
			do { 
				ret = ret * 10 + c - '0'; 
			} 
			while ((c = read()) >= '0' && c <= '9'); 
			if (neg) 
				return -ret; 
			return ret; 
		} 
  
		public double nextDouble() throws IOException 
		{ 
			double ret = 0, div = 1; 
			byte c = read(); 
			while (c <= ' ') 
				c = read(); 
			boolean neg = (c == '-'); 
			if (neg) 
				c = read(); 
  
			do { 
				ret = ret * 10 + c - '0'; 
			} 
			while ((c = read()) >= '0' && c <= '9'); 
  
			if (c == '.') 
			{ 
				while ((c = read()) >= '0' && c <= '9') 
				{ 
					ret += (c - '0') / (div *= 10); 
				} 
			} 
  
			if (neg) 
				return -ret; 
			return ret; 
		} 
		
		
  
		private void fillBuffer() throws IOException 
		{ 
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
			if (bytesRead == -1) 
				buffer[0] = -1; 
		} 
  
		private byte read() throws IOException 
		{ 
			if (bufferPointer == bytesRead) 
				fillBuffer(); 
			return buffer[bufferPointer++]; 
		} 
  
		public void close() throws IOException 
		{ 
			if (din == null) 
				return; 
			din.close(); 
		} 
	} 
}