A. Common Prefixes

A. Common Prefixes

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The length of the longest common prefix of two strings s=s1s2…sns=s1s2…sn and t=t1t2…tmt=t1t2…tm is defined as the maximum integer kk (0≤k≤min(n,m)0≤k≤min(n,m)) such that s1s2…sks1s2…sk equals t1t2…tkt1t2…tk.

Koa the Koala initially has n+1n+1 strings s1,s2,…,sn+1s1,s2,…,sn+1.

For each ii (1≤i≤n1≤i≤n) she calculated aiai — the length of the longest common prefix of sisi and si+1si+1.

Several days later Koa found these numbers, but she couldn't remember the strings.

So Koa would like to find some strings s1,s2,…,sn+1s1,s2,…,sn+1 which would have generated numbers a1,a2,…,ana1,a2,…,an. 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 tt (1≤t≤1001≤t≤100) — the number of test cases. Description of the test cases follows.

The first line of each test case contains a single integer nn (1≤n≤1001≤n≤100) — the number of elements in the list aa.

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤500≤ai≤50) — the elements of aa.

It is guaranteed that the sum of nn over all test cases does not exceed 100100.

Output

For each test case:

Output n+1n+1 lines. In the ii-th line print string sisi (1≤|si|≤2001≤|si|≤200), consisting of lowercase Latin letters. Length of the longest common prefix of strings sisi and si+1si+1 has to be equal to aiai.

If there are many answers print any. We can show that answer always exists for the given constraints.

Example

input

Copy

4
4
1 2 4 2
2
5 3
3
1 3 1
3
0 0 0

output

Copy

aeren
ari
arousal
around
ari
monogon
monogamy
monthly
kevinvu
kuroni
kurioni
korone
anton
loves
adhoc
problems

Note

In the 11-st test case one of the possible answers is s=[aeren,ari,arousal,around,ari]s=[aeren,ari,arousal,around,ari].

Lengths of longest common prefixes are:

• Between aerenaeren and ariari →1→1
• Between ariari and arousalarousal →2→2
• Between arousalarousal and aroundaround →4→4
• Between aroundaround and ariari →2
• ======================================================================
• 把第一个构造成全是a,其余直接根据a[i]构造即可，保证前缀相等的同时，其余全部搞成相反的(a->b,b->a)，长度最小是1

  # include<iostream>
  # include<string.h>
  # include<vector>
  
  using namespace std;
  
  int a[1000];
  string s[1000];
  
  int main ()
  {
  
     int t;
     cin>>t;
  
     while(t--)
     {
         int n;
         cin>>n;
         int maxx=1;
         for(int i=1;i<=n;i++)
         {
             cin>>a[i];
             maxx=max(maxx,a[i]);
             s[i]="";
         }
         s[n+1]="";
  
         for(int i=1;i<=maxx;i++)
         {
             s[1]+='a';
  
         }
  
         for(int i=2;i<=n+1;i++)
         {
             for(int j=0;j<a[i-1];j++)
             {
                 s[i]+=s[i-1][j];
             }
             for(int j=a[i-1];j<maxx;j++)
             {
                 if(s[i-1][j]=='a')
                  s[i]+='b';
                 else
                  s[i]+='a';
             }
  
         }
  
         for(int i=1;i<=n+1;i++)
         {
             cout<<s[i]<<endl;
         }
  
     }
  
      return  0;
  
  }