An encoder encodes the first $16$ lowercase English letters using $4$ bits each. The first bit (from the left) of the code is $0$ if the letter lies among the first $8$ letters, else it is $1$, signifying that it lies among the last $8$ letters. The second bit of the code is $0$ if the letter lies among the first $4$ letters of those $8$ letters found in the previous step, else it's $1$, signifying that it lies among the last $4$ letters of those $8$ letters. Similarly, the third and the fourth bit each signify the half in which the letter lies.
For example, the letter $j$ would be encoded as :
- Among $(a,b,c,d,e,f,g,h$ $|$ $i,j,k,l,m,n,o,p)$, $j$ appears in the second half. So the first bit of its encoding is $1$.
- Now, among $(i,j,k,l$ $|$ $m,n,o,p)$, $j$ appears in the first half. So the second bit of its encoding is $0$.
- Now, among $(i,j$ $|$ $k,l)$, $j$ appears in the first half. So the third bit of its encoding is $0$.
- Now, among $(i$ $|$ $j)$, $j$ appears in the second half. So the fourth and last bit of its encoding is $1$.
So $j$'s encoding is $1001$,
Given a binary encoded string $S$, of length at most $10^5$, decode the string. That is, the first 4 bits are the encoding of the first letter of the secret message, the next 4 bits encode the second letter, and so on. It is guaranteed that the string's length is a multiple of 4.

-----Input:-----
- The first line of the input contains an integer $T$, denoting the number of test cases.
- The first line of each test case contains an integer $N$, the length of the encoded string.
- The second line of each test case contains the encoded string $S$.

-----Output:-----
For each test case, print the decoded string, in a separate line.

-----Constraints-----
- $1 \leq T \leq 10$
- $4 \leq N \leq 10^5$
- The length of the encoded string is a multiple of $4$.
- $0 \le S_i \le 1$

-----Subtasks-----
- $100$ points : Original constraints.

-----Sample Input:-----
3
4
0000
8
00001111
4
1001

-----Sample Output:-----
a
ap
j

-----Explanation:-----
- Sample Case $1$ :
The first bit is $0$, so the letter lies among the first $8$ letters, i.e., among $a,b,c,d,e,f,g,h$. The second bit is $0$, so it lies among the first four of these, i.e., among $a,b,c,d$.
The third bit is $0$, so it again lies in the first half, i.e., it's either $a$ or $b$. Finally, the fourth bit is also $0$, so we know that the letter is $a$.
- Sample Case $2$ :
Each four bits correspond to a character. Just like in sample case $1$, $0000$ is equivalent to $a$. Similarly, $1111$ is equivalent to $p$. So, the decoded string is $ap$.
- Sample Case $3$ :
The first bit is $1$, so the letter lies among the last $8$ letters, i.e., among $i,j,k,l,m,n,o,p$. The second bit is $0$, so it lies among the first four of these, i.e., among $i,j,k,l$.
The third bit is $0$, so it again lies in the first half, i.e., it's either $i$ or $j$. Finally, the fourth bit is $1$, so we know that the letter is $j$.

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