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 org.w3c.dom.ls.LSOutput;

import java.util.*;
import java.util.function.IntBinaryOperator;


public class Example {
    static long[][] dp;
    //  static    int[]  ar;
    static int mod = 1000000000 + 7;
    static long[] ans = new long[100000 + 1];
    static List<List<Integer>> adj;
    static int[] vis;
    static int[] dis;
    static int row;
    static boolean[] chosen;
    static List<Integer> permutation;
    static int k;
    static int count;
    static int[][] arr;

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
         int t=sc.nextInt();
         while(t>0){
             t--;
             int n = sc.nextInt();
             int [] arr= new int[n];
             char[] cc= new char[100];
             for(int i=0;i<100;i++){
                 cc[i]='a';
             }
             System.out.println(String.valueOf(cc));
             for(int i=0;i<n;i++){
                 int a= sc.nextInt();
                 cc[a]=cc[a]=='a'?'b':'a';
                 System.out.println(String.valueOf(cc));

             }





         }
    }

    private static long getans1(int[] s, int[] c, int i, int cc, int maxValue) {


        if (cc == 1 && i >= 0) {
            return c[i];
        } else if (cc == 0 && i >= 0) {
            return 0;
        } else if (i < 0) {
            return Integer.MAX_VALUE;
        }
        if (dp[i][cc] != -1) {
            return dp[i][cc];
        }
        if (s[i] < maxValue || maxValue == 0) {
            return dp[i][cc] = Math.min(getans1(s, c, i - 1, cc, maxValue), c[i] + getans1(s, c, i - 1, cc - 1, s[i]));

        } else {
            return dp[i][cc] = getans1(s, c, i - 1, cc, maxValue);

        }
    }


    private static void swap(int i, int j, int[] arrr) {
        int temp = arrr[i];
        arrr[i] = arrr[j];
        arrr[j] = temp;
    }

    private static long getsum(int[][] h, int i, int p, long[][] dp) {

        if (i >= h[0].length) {
            return 0;
        }
        if (dp[i][p] != -1) {
            return dp[i][p];
        }

        if (p == 0) {
            long a = getsum(h, i + 1, p, dp);
            a = Math.max(a, h[1][i] + getsum(h, i + 1, 1, dp));
            a = Math.max(a, h[2][i] + getsum(h, i + 1, 2, dp));
            return dp[i][p] = a;
        } else if (p == 1) {
            long a = getsum(h, i + 1, p, dp);
            //a=Math.max(a,h[i][1]+getsum(h,i+1,1));
            a = Math.max(a, h[2][i] + getsum(h, i + 1, 2, dp));
            return dp[i][p] = a;
        } else {
            long a = getsum(h, i + 1, p, dp);
            a = Math.max(a, h[1][i] + getsum(h, i + 1, 1, dp));
            //  a=  Math.max(a,h[i][2]+getsum(h,i+1,2));
            return dp[i][p] = a;
        }

    }

    private static int getans(int[] arr, int i) {
        if (i < 0)
            return 0;
        else {
            int min = mod;
            for (int j = 0; j < arr.length; j++) {
                if (arr[j] == 1) {
                    arr[j] = 2;
                    for (int k = 0; k < j; k++) {
                        if (arr[k] == 0) {
                            arr[k] = 2;
                            min = Math.min(min, Math.abs(j - k) + getans(arr, i - 1));
                        }
                    }
                }
            }
            return min;
        }
    }

    private static int ncr(int notsure, int i) {
        int a = 1;
        a = (factorail(notsure)) / (factorail(notsure - 1) * (factorail(i)));
        return a;
    }

    private static int factorail(int notsure) {
        int fac = 1;
        for (int i = 2; i <= notsure; i++) {
            fac = fac + i;
        }
        return fac;

    }

    private static void allPerMutation(int n) {
        if (permutation.size() == n) {
            //   System.out.println(permutation);
            long sum = arr[1][permutation.get(0)];
            for (int i = 0; i < permutation.size() - 1; i++) {
                sum += arr[permutation.get(i)][permutation.get(i + 1)];
            }
            sum = sum + arr[permutation.get(n - 1)][1];
            if (sum == k) {
                count++;
            }

        } else {
            for (int i = 2; i <= (n + 1); i++) {
                if (chosen[i]) {
                    continue;
                }
                permutation.add(i);
                chosen[i] = true;
                allPerMutation(n);
                permutation.remove(permutation.size() - 1);
                chosen[i] = false;
            }


        }


    }


    public static int Check(int[] arr, int k) {
        int sum = 0;
        int j = 1;
        int total = 0;
        for (int i = 1; i < arr.length; i++) {
            sum += arr[i];
            if (sum == k) {
                total += (i - j);
                j = i + 1;
                sum = 0;
            }
            if (sum > k) {
                return arr.length;
            }
        }
        return total;


    }

    public static char[][] rotateTheBox(char[][] box) {
        Stack<Character> st;
        int m = box.length;
        int n = box[0].length;
        char[][] cc = new char[box[0].length][box.length];
        int k = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                cc[i][j] = '.';
            }
        }
        for (int i = m - 1; i >= 0; i--) {

            st = new Stack<>();
            for (int j = 0; j < n; j++) {
                char c = box[i][j];
                if (c == '#') {
                    st.push(c);
                } else if (c == '*') {
                    cc[j][k] = '*';
                    for (int p = j - 1; p >= 0; p--) {
                        if (!st.isEmpty())
                            cc[p][k] = st.pop();
                        else
                            break;
                    }
                    st = new Stack<>();
                }
            }
            if (!st.isEmpty()) {
                for (int p = n - 1; p >= 0; p--) {
                    if (!st.isEmpty())
                        cc[p][k] = st.pop();
                    else
                        break;
                }
            }
            k++;


        }
        return cc;


    }


    private static long findans(int[] arr, int i, int s) {
        if (i < 0 && s == 0)
            return 0;
        if (i < 0 && s == 1)
            return 0;
        if (dp[i][s] != -1) {
            return dp[i][s];
        } else {


            if (s == 0)
                return dp[i][s] = Math.max(findans(arr, i - 1, s), -1 * arr[i] + findans(arr, i - 1, 1 - s));
            else
                return dp[i][s] = Math.max(findans(arr, i - 1, s), arr[i] + findans(arr, i - 1, 1 - s));
        }
    }

    public static void dfs(int v, int di) {

        for (int i = 0; i < adj.get(v).size(); i++) {
            int x = adj.get(v).get(i);
            if (dis[x] > (di + 1)) {
                dis[x] = di + 1;
                dfs(x, di + 1);
            }
        }
    }

    private static boolean isPrime(long a) {
        if (a == 1)
            return true;

        for (int i = 2; i * i <= a; i++) {
            if (a % i == 0)
                return false;

        }
        return true;
    }

    private static int gcd(int aa, int bb) {
        if (bb == 0)
            return aa;
        return gcd(bb, aa % bb);

    }

    private static int lowerBound(List<Integer> ar, int a) {
        int res = -1;
        int s = 0;
        int e = ar.size() - 1;
        while (s <= e) {
            int mid = (s) + (e - s) / 2;
            if (ar.get(mid) <= a) {
                res = mid;
                s = mid + 1;

            } else {
                e = mid - 1;
            }
        }

        return res;
    }

    private static int upperBound(List<Integer> ar, int a) {
        int res = -1;
        int s = 0;
        int e = ar.size() - 1;
        while (s <= e) {
            int mid = (s) + (e - s) / 2;
            if (ar.get(mid) <= a) {

                s = mid + 1;

            } else {
                res = mid;
                e = mid - 1;
            }
        }

        return res;
    }

    public static int getXORSum(int[] arr1, int[] arr2) {

        long c = 0;
        int n = arr1.length;
        int m = arr2.length;
        if (n == 1 && m == 1) {
            return arr1[0] & arr2[0];
        }
        int aa = n + m;
        for (int i = 0; i < 32; i++) {

            int z = 0;
            int o = 0;
            int cc = 0;
            for (int a : arr1) {


                int x = (a & (1 << i));
                if (x == 0)
                    z++;
                else
                    o++;


            }
            for (int a : arr2) {
                int x = (a & (1 << i));
                if (x == 0)
                    z++;
                else
                    o++;
            }

            System.out.println(z + " " + o);

            if (z != aa && o != aa) {
                c = c + (1 << i);

            }


        }
        return (int) c;


    }

    static boolean[] dpbo;

    public static boolean canReach(String s, int minJump, int maxJump, int i) {
        if (i == 0) {
            return true;
        }
        if (i < 0) {
            return false;
        }

        boolean tt = false;
        for (int j = minJump; j <= maxJump; j++) {
            if (i - j >= 0 && s.charAt(i - j) == '0') {
                tt = (tt | canReach(s, minJump, maxJump, i - j));
            }

        }
        return tt;


    }
    public static boolean canReach(String s, int minJump, int maxJump) {

        adj=new ArrayList<>();
        int n = s.length();
        for(int i=0;i<n;i++){
            adj.add(new ArrayList<>());
        }
        // int c=0;
        vis=new int[n];
        for(int i=0;i<n;i++){
            if(s.charAt(i)=='0'){
              //  System.out.println(i+" ");


                for(int j=i+minJump;j<=i+maxJump;j++){

                    if(j<n){
                        if(s.charAt(j)=='0'){
                         //   System.out.println("HELOO");
                            adj.get(i).add(j);

                        }
                    }
                }


            }
        }
        int c=0;
        dfs(0);

        return  vis[n-1] == 1;


    }
    public static void dfs(int v){


        vis[v]=1;
        for(int i=0;i<adj.get(v).size();i++){
            int x=adj.get(v).get(i);
            if(vis[x]==0){
                dfs(x);
            }
        }









    }


}