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:
//package Codeforce.div2.div2_659;

import java.io.*;
import java.util.StringTokenizer;

/**
 * BEGIN NO SAD
 * Blessed are those who suffer for doing what is right.
 * The kingdom of heaven belongs to them.
 * (Matthew 5:10-12)
 * @author Tran Anh Tai
 * @template for CP codes
 * END NO SAD
 */
public class TaskA {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }
    // main solver
    static class Task{
        public void solve(InputReader in, PrintWriter out) {
            int test = Integer.parseInt(in.nextToken());
            for (int i = 0; i < test; i++) {
                int n = in.nextInt();
                int[]a = new int[n];
                for (int j = 0; j < n; j++) {
                    a[j] = in.nextInt();
                }
                String tar = gen('a', 50);
                for (int j = 0; j < n; j++){
                    out.println(tar);
                    char c = 'a';
                    if (a[j] < tar.length() && tar.charAt(a[j]) == 'a'){
                        c = 'b';
                    }
                    tar = tar.substring(0, a[j]) + gen(c, 50 - a[j]);
                }
                out.println(tar);
            }
        }
        public String gen(char ch, int t){
            String r = "";
            for(int i= 0;i  < t; i++)r += ch;
            return r;
        }
    }
    static class Point{
        public double x, y;
        public Point(double x, double y){
            this.x = x; this.y = y;
        }
        public static Line getBisect(Point A, Point B){
            double a = B.x - A.x;
            double b = B.y - A.y;
            double c = a * (A.x + B.x) / 2 + b * (A.y + B.y) / 2;
            return new Line(a, b, c);
        }
        public static Point center(Point A, Point B, Point C){
            Line lab = getBisect(A, B);
            Line lac = getBisect(A, C);
            Point intersect = Line.intersect(lab, lac);
            return intersect;
        }
    }
    static class Line{
        public double a, b, c;
        public Line(double a, double b, double c){
            this.a = a; this.b = b; this.c = c;
        }
        public static Point intersect(Line l1, Line l2){
            double D = l1.a * l2.b - l1.b * l2.a;
            double Dy = l1.a * l2.c - l2.a * l1.c;
            double Dx = l1.c * l2.b - l2.c * l1.b;
            return new Point(Dx / D, Dy / D);
        }
    }
    // fast input reader class;
    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                String line = null;
                try {
                    line = br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (line == null) {
                    return null;
                }
                st = new StringTokenizer(line);
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        public long nextLong(){
            return Long.parseLong(nextToken());
        }
    }
}