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.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Round659 {

	
	static int X[] = { -1, 0, 0, 1 };
	static int Y[] = { 0, -1, 1, 0 };

	static class Reader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;

		public Reader() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public Reader(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();
		}
	}

	public static long[] initArray(int n, Reader scan) throws IOException {

		long arr[] = new long[n];
		for (int i = 0; i < n; i++) {
			arr[i] = scan.nextLong();
		}

		return arr;
	}

	public static long sum(long arr[]) {

		long sum = 0;
		for (long i : arr) {
			sum += (long) i;
		}

		return sum;
	}

	public static long max(long arr[]) {

		long max = Long.MIN_VALUE;

		for (long i : arr) {
			max = Math.max(i, max);
		}

		return max;
	}

	public static long min(long arr[]) {
		long min = Long.MAX_VALUE;

		for (long i : arr) {
			min = Math.min(i, min);
		}

		return min;
	}

	public static List<Integer>[] initAdjacency(int n, int e, Reader scan, boolean type) throws IOException {

		List<Integer> adj[] = new ArrayList[n + 1];

		for (int i = 0; i < e; i++) {
			int u = scan.nextInt();
			int v = scan.nextInt();

			if (adj[u] == null)
				adj[u] = new ArrayList<>();
			if (type && adj[v] == null)
				adj[v] = new ArrayList<>();

			adj[u].add(v);

			if (type)
				adj[v].add(u);

		}

		return adj;
	}
	public static void main(String[] args) throws IOException {
		
//		Reader scan=new Reader();
		Scanner scan=new Scanner(System.in);
		int t=scan.nextInt();
		while(t-->0) {
			int n=scan.nextInt();
			int arr[]=new int[n];
			
			for(int i=0;i<n;i++) {
				arr[i]=scan.nextInt();
			}
			A(n, arr);
			
		}

	
	}
	
	public static void A(int n, int arr[]) {
		
		StringBuilder ans[]=new StringBuilder[n+1];
		for(int i=0;i<ans.length;i++) {
			ans[i]=new StringBuilder("");	
		}
		
		for(int i=0;i<2;i++) {
			for(int j=0;j<26;j++) {
				ans[0].append((char)(j+'a'));
			}
		}
		
		for(int i=0;i<arr.length;i++) {
			
			String last= ans[i].toString();
			
			for(int j=0;j<arr[i];j++) {
				ans[i+1].append(last.charAt(j));
			}
			
			char nextChar= last.charAt(arr[i]);
			
			for(int j=arr[i]+1;j<last.length();j++) {
				ans[i+1].append(last.charAt(j));
			}
			
			ans[i+1].append(nextChar);
			
		}
		
		for(StringBuilder s: ans) {
			System.out.println(s.toString());
		}
		
	}
	
	public static void C(int n, String a, String b) {
		
		int count[][]=new int[20][20];
		
		for(int i=0;i<a.length();i++) {
			
			
			
			if(a.charAt(i)>b.charAt(i)) {
				System.out.println(-1);
				return;
			}else if(a.charAt(i)<b.charAt(i)){
				count[a.charAt(i)-'a'][b.charAt(i)-'a']++;
			}
			
		}
		
		
		int ans=0;
		outer: for(int i=0;i<20;i++) {
			for(int j=0;j<20;j++) {
				if(count[i][j]!=0) {
					ans++;
					
					for(int k=j+1;k<20;k++) {
						count[j][k]+= count[i][k];
					}
					continue outer;
					
				}
			}
		}
		
		System.out.println(ans);
		
	}
}