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.lang.*;
import java.math.*;
import java.io.*;
import java.util.HashSet;
import java.util.Arrays; 
import java.util.Scanner;
import java.util.Set;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.text.DecimalFormat;
import java.lang.Math;
import java.lang.reflect.Array;
import java.util.Iterator; 
import java.util.TreeSet;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigDecimal;
import java.util.*;
public class Codeforces{
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	static long MOD = (long)(1e9+7);
	static FastReader sc = new FastReader();
	static int pInf = Integer.MAX_VALUE;
	static int nInf = Integer.MIN_VALUE;
	public static void main(String[] args){
		int test = 1;
		test = sc.nextInt();
		while(test-->0){
		    int n = sc.nextInt();
		    char[][] c = new char[n+1][51];
		    for(int i = 0; i < n+1; i++) {
		    	for(int j = 0; j < 51; j++) {
		    		c[i][j] = 'a';
		    	}
		    }
		    for(int i = 1; i < n+1; i++) {
		    	int x = sc.nextInt();
		    	for(int j = 0; j < x; j++) {
		    		c[i][j] = c[i-1][j];
		    	}
		    	c[i][x]  = c[i-1][x]=='a'?'b':'a';
		    }
		    for(int i = 0; i < n+1; i++) {
		    	out.println(new String(c[i]));
		    }
		}
		out.close();
	}
	public static long mul(long a, long b){
	    return ((a%MOD)*(b%MOD))%MOD;
	}
	public static long add(long a, long b){
	    return ((a%MOD)+(b%MOD))%MOD;
	}
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//Integer.lowestOneBit(i)   Equals k where k is the position of the first one in the binary
//Integer.highestOneBit(i)  Equals k where k is the position of the last one in the binary
//Integer.bitCount(i)       returns the number of one-bits
//Collections.sort(A,(p1,p2)->(int)(p2.x-p1.x)) To sort ArrayList in descending order wrt values of x.
//            Arrays.parallelSort(a,new Comparator<TPair>() {
//      		public int compare(TPair a,TPair b) {
//        			if(a.y==b.y) return a.x-b.x;
//        			return b.y-a.y;
//        		}
//        	});
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//PrimeFactors    
    public static ArrayList<Long> primeFactors(long n) {
        ArrayList<Long> arr = new ArrayList<>();
        if (n % 2 == 0)
            arr.add((long) 2);
        while (n % 2 == 0)
            n /= 2;
        for (long i = 3; i <= Math.sqrt(n); i += 2) {
            int flag = 0;
            while (n % i == 0) {
                n /= i;
                flag = 1;
            }
            if (flag == 1)
                arr.add(i);
        }
        if (n > 2)
            arr.add(n);
        return arr;
    }
//Pair Class
	static class Pair implements Comparable<Pair>{
		int x;
		int y;
		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
		@Override
		public int compareTo(Pair o) {
			if(this.x==o.x){
			    return (this.y-o.y);
			}
			return (this.x-o.x);
		}
	}
	static class TPair{
		long x;
		long y;
		public TPair(long x, long y) {
			this.x = x;
			this.y = y;
		}
	}
//Merge Sort
	static void merge(long arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        long L[] = new long [n1];
        long R[] = new long [n2];
 
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
    static void sort(long arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
//Brian Kernighan’s Algorithm
    static long countSetBits(long n){
        if(n==0) return 0;
        return 1+countSetBits(n&(n-1));
    }
//Factorial
    static long factorial(long n){
        if(n==1) return 1;
        if(n==2) return 2;
        if(n==3) return 6;
        return n*factorial(n-1);
    }
//Euclidean Algorithm
    static long gcd(long A,long B){
        if(B==0) return A;
        return gcd(B,A%B);
    }
//Modular Exponentiation
    static long fastExpo(long x,long n){
        if(n==0) return 1;
        if((n&1)==0) return fastExpo((x*x)%MOD,n/2)%MOD;
        return ((x%MOD)*fastExpo((x*x)%MOD,(n-1)/2))%MOD;
    }
//AKS Algorithm
    static boolean isPrime(long n){
        if(n<=1) return false;
        if(n<=3) return true;
        if(n%2==0 || n%3==0) return false;
        for(int i=5;i<=Math.sqrt(n);i+=6)
            if(n%i==0 || n%(i+2)==0) return false;
        return true;
    }
//Reverse an array
    static <T> void reverse(T arr[],int l,int r){
    	Collections.reverse(Arrays.asList(arr).subList(l, r));
    }
//Sieve of eratosthenes
    static int[] findPrimes(int n){
        boolean isPrime[]=new boolean[n+1];
        ArrayList<Integer> a=new ArrayList<>();
        int result[];
        Arrays.fill(isPrime,true);
        isPrime[0]=false;
        isPrime[1]=false;
        for(int i=2;i*i<=n;++i){
            if(isPrime[i]==true){
                for(int j=i*i;j<=n;j+=i) isPrime[j]=false;
            }
        }
        for(int i=0;i<=n;i++) if(isPrime[i]==true) a.add(i);
        result=new int[a.size()];
        for(int i=0;i<a.size();i++) result[i]=a.get(i);
        return result;
        
    }
//Segmented Sieve
    static boolean[] segmentedSieve(long l, long r){
    	boolean[] segSieve = new boolean[(int)(r-l+1)];
    	Arrays.fill(segSieve, true);
    	int[] prePrimes = findPrimes((int)Math.sqrt(r));
    	for(int p:prePrimes) {
    		long low = (l/p)*p;
    		if(low < l) {
    			low += p;
    		}
    		if(low == p) {
    			low += p;
    		}
    		for(long j = low; j<= r; j += p) {
    			segSieve[(int) (j-l)] = false;
    		}
    	}
    	if(l==1) {
    		segSieve[0] = false;
    	}
    	return segSieve;
    }
//Euler Totent function
    static long countCoprimes(long n){
        ArrayList<Long> prime_factors=new ArrayList<>();
        long x=n,flag=0;
        while(x%2==0){
            if(flag==0) prime_factors.add(2L);
            flag=1;
            x/=2;
        }
        for(long i=3;i*i<=x;i+=2){
            flag=0;
            while(x%i==0){
                if(flag==0) prime_factors.add(i);
                flag=1;
                x/=i;
            }
        }
        if(x>2) prime_factors.add(x);
        double ans=(double)n;
        for(Long p:prime_factors){
            ans*=(1.0-(Double)1.0/p);
        }
        return (long)ans;
    }
	static long modulo = (long)1e9+7;
	public static long modinv(long x){
	    return modpow(x, modulo-2);
	}
	public static long modpow(long a, long b){
	    if(b==0){
	        return 1;
	    }
	    long x = modpow(a, b/2);
	    x = (x*x)%modulo;
	    if(b%2==1){
	        return (x*a)%modulo;
	    }
	    return x;
	}
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception r) {
					r.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception r) {
				r.printStackTrace();
			}
			return str;
		}
	}
}