Polycarp has come up with a new game to play with you. He calls it "A missing bigram".

A bigram of a word is a sequence of two adjacent letters in it.

For example, word "abbaaba" contains bigrams "ab", "bb", "ba", "aa", "ab" and "ba".

The game goes as follows. First, Polycarp comes up with a word, consisting only of lowercase letters 'a' and 'b'. Then, he writes down all its bigrams on a whiteboard in the same order as they appear in the word. After that, he wipes one of them off the whiteboard.

Finally, Polycarp invites you to guess what the word that he has come up with was.

Your goal is to find any word such that it's possible to write down all its bigrams and remove one of them, so that the resulting sequence of bigrams is the same as the one Polycarp ended up with.

The tests are generated in such a way that the answer exists. If there are multiple answers, you can print any of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of testcases.

The first line of each testcase contains a single integer n (3 ≤ n ≤ 100) — the length of the word Polycarp has come up with.

The second line of each testcase contains n-2 bigrams of that word, separated by a single space. Each bigram consists of two letters, each of them is either 'a' or 'b'.

Additional constraint on the input: there exists at least one string such that it is possible to write down all its bigrams, except one, so that the resulting sequence is the same as the sequence in the input. In other words, the answer exists.

Output

For each testcase print a word, consisting of n letters, each of them should be either 'a' or 'b'. It should be possible to write down all its bigrams and remove one of them, so that the resulting sequence of bigrams is the same as the one Polycarp ended up with.

The tests are generated in such a way that the answer exists. If there are multiple answers, you can print any of them. 

Example

Input


4
7
ab bb ba aa ba
7
ab ba aa ab ba
3
aa
5
bb ab bb


Output


abbaaba
abaabaa
baa
bbabb

Note

The first two testcases from the example are produced from the word "abbaaba". As listed in the statement, it contains bigrams "ab", "bb", "ba", "aa", "ab" and "ba".

In the first testcase, the 5-th bigram is removed. 

In the second testcase, the 2-nd bigram is removed. However, that sequence could also have been produced from the word "abaabaa". It contains bigrams "ab", "ba", "aa", "ab", "ba" and "aa". The missing bigram is the 6-th one.

In the third testcase, all of "baa", "aab" and "aaa" are valid answers.
4
7
ab bb ba aa ba
7
ab ba aa ab ba
3
aa
5
bb ab bb

abbaaba
abaabaa
aaa
bbabb

from __future__ import division, print_function

import os,sys
from io import BytesIO, IOBase

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

from bisect import bisect_left as lower_bound, bisect_right as upper_bound 
def so():  return int(input())
def st():  return input()
def mj():  return map(int,input().strip().split(" "))
def msj(): return list(map(str,input().strip().split(" ")))
def le():  return list(map(int,input().split()))
def rc():  return map(float,input().split())
def lebe():return list(map(int, input()))

def dmain():
    sys.setrecursionlimit(1000000)
    threading.stack_size(1024000)
    thread = threading.Thread(target=main)
    thread.start()
def joro(L):
    return(''.join(map(str, L)))


def decimalToBinary(n): return bin(n).replace("0b","")


def isprime(n):
    for i in range(2,int(n**0.5)+1):
        if n%i==0:
            return False
    return True
def npr(n, r):
    return factorial(n) // factorial(n - r) if n >= r else 0
 
 
def ncr(n, r):
    return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0
 
 
def lower_bound(li, num):
    answer = -1
    start = 0
    end = len(li) - 1
 
    while (start <= end):
        middle = (end + start) // 2
        if li[middle] >= num:
            answer = middle
            end = middle - 1
        else:
            start = middle + 1
    return answer  # min index where x is not less than num
 
 
def upper_bound(li, num):
    answer = -1
    start = 0
    end = len(li) - 1
 
    while (start <= end):
        middle = (end + start) // 2
 
        if li[middle] <= num:
            answer = middle
            start = middle + 1
 
        else:
            end = middle - 1
    return answer  # max index where x is not greater than num
def tir(a,b,c):
    if(0==c):
        return 1
    if(len(a)<=b):
        return 0
    
    if(c!=-1):
        return (tir(a,1+b,c+a[b]) or tir(a,b+1,c-a[b]) or tir(a,1+b,c)) 
        
        
    else:
        return (tir(a,1+b,a[b]) or tir(a,b+1,-a[b]) or tir(a,1+b,-1))
hoi=int(2**20)        
 
def abs(x):
    return x if x >= 0 else -x
 
 
def binary_search(li, val, lb, ub):
    # print(lb, ub, li)
    ans = -1
    while (lb <= ub):
        mid = (lb + ub) // 2
        # print('mid is',mid, li[mid])
        if li[mid] > val:
            ub = mid - 1
        elif val > li[mid]:
            lb = mid + 1
        else:
            ans = mid  # return index
            break
    return ans
 
 
def kadane(x):  # maximum sum contiguous subarray
    sum_so_far = 0
    current_sum = 0
    for i in x:
        current_sum += i
        if current_sum < 0:
            current_sum = 0
        else:
            sum_so_far = max(sum_so_far, current_sum)
    return sum_so_far
 
 
def pref(li):
    pref_sum = [0]
    for i in li:
        pref_sum.append(pref_sum[-1] + i)
    return pref_sum
 
 
def SieveOfEratosthenes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    li = []
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1
 
    for p in range(2, len(prime)):
        if prime[p]:
            li.append(p)
    return li
 
 
def primefactors(n):
    factors = []
    while (n % 2 == 0):
        factors.append(2)
        n //= 2
    for i in range(3, int(sqrt(n)) + 1, 2):  # only odd factors left
        while n % i == 0:
            factors.append(i)
            n //= i
    if n > 2:  # incase of prime
        factors.append(n)
    return factors
         
    
def read():
    sys.stdin  = open('input.txt', 'r')  
    sys.stdout = open('output.txt', 'w') 
def tr(n):
    return n*(n+1)//2
boi=int(1e9-7)
doi=int(1e9+7)

        
    
def iu():
    import sys
    import math as my
    input=sys.stdin.readline
    m=so()
    L=msj()
    s=""
    o=0
    for i in range(len(L)):
        if(i==0):
            s+=L[i]
        elif(L[i][0]==s[len(s)-1]):
            s+=L[i][1]
        else:
            s+=L[i]
            o=1
    if(o==0):
        print(s+"b")
    else:
        print(s)
        
        
    
        
    
    
    
def main():
    for i in range(so()):
        iu()
        
    
    
    
    
    
    
    
    
    
    

    

    
        
        
            
    
    
                
            
                
                
                
                
            
    
            
            
    
    
    
    
        
            
    
           
            
            
        
    
        
    
    
        
        
        

            
                    
                
                    
                    
                    
                    
                    
                    
        
                
        
        
       
    
           
          
          
                
            
        
                
    
        
        
        
       
            
                
        




# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion


if __name__ == "__main__":
    #read()
    main()
    #dmain()

# Comment Read()