Problem: You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n=3 but there is 4 in the array).

You can perform at most n-1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.

Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.

You can see the definition of the lexicographical order in the notes section.

You have to answer q independent test cases.

For example, let's consider the permutation [5, 4, 1, 3, 2]. The minimum possible permutation we can obtain is [1, 5, 2, 4, 3] and we can do it in the following way:

  1. perform the second operation (swap the second and the third elements) and obtain the permutation [5, 1, 4, 3, 2]; 
  2. perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5, 1, 4, 2, 3]; 
  3. perform the third operation (swap the third and the fourth elements) and obtain the permutation [5, 1, 2, 4, 3]. 
  4. perform the first operation (swap the first and the second elements) and obtain the permutation [1, 5, 2, 4, 3]; 



Another example is [1, 2, 4, 3]. The minimum possible permutation we can obtain is [1, 2, 3, 4] by performing the third operation (swap the third and the fourth elements).

Input

The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of test cases. Then q test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 100) — the number of elements in the permutation.

The second line of the test case contains n distinct integers from 1 to n — the given permutation.

Output

For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.

Example

Input


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


Output


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

Note

Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i ≤ n that for all j from 1 to i - 1 the condition p_j = q_j is satisfied, and p_i < q_i. For example:

  * p = [1, 3, 5, 2, 4] is less than q = [1, 3, 5, 4, 2] (such i=4 exists, that p_i < q_i and for each j < i holds p_j = q_j), 
  * p = [1, 2] is less than q = [2, 1] (such i=1 exists, that p_i < q_i and for each j < i holds p_j = q_j). 

Solution (in PYTHON3):

Solution:
#from collections import OrderedDict
#from collections import defaultdict
#from functools import reduce
#from itertools import groupby
#sys.setrecursionlimit(10**6)
#from itertools import accumulate
from collections import Counter
import math
import sys
import os
from io import BytesIO, IOBase
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")
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
'''
from threading import stack_size,Thread
sys.setrecursionlimit(10**6)
stack_size(10**8)
'''
def listt():
    return [int(i) for i in input().split()]
 
def gcd(a,b): 
   return math.gcd(a,b) 
 
def lcm(a,b): 
    return (a*b) // gcd(a,b) 

def factors(n):
    step = 2 if n%2 else 1
    return set(reduce(list.__add__,([i, n//i] for i in range(1, int(math.sqrt(n))+1, step) if n % i == 0)))

def comb(n,k):
    factn=math.factorial(n)
    factk=math.factorial(k)
    fact=math.factorial(n-k)
    ans=factn//(factk*fact)
    return ans

def is_prime(n): 
    if n <= 1: 
        return False
    if n == 2: 
        return True
    if n > 2 and n % 2 == 0: 
        return False

    max_div = math.floor(math.sqrt(n)) 
    for i in range(3, 1 + max_div, 2): 
        if n % i == 0: 
            return False
    return True
def maxpower(n,x):
    B_max = int(math.log(n, x)) + 1 #tells upto what power of x n is less than it like 1024->5^4
    return B_max
def binaryToDecimal(n): 
    return int(n,2) 

#d=sorted(d.items(),key=lambda a:a[1]) sort dic accc to values
#g=pow(2,6)
#a,b,n,m=map(int,input().split())
#print ("{0:.8f}".format(min(l)))
#t = int(input())
t=int(input())
for _ in range(t):
    res = []
    n = int(input())
    a = listt()
    l = 0
    for m in range(1, n+1):
        i = a.index(m)
        if i <= l:
            continue
        for j in range(i-1, l-1, -1):
            if a[j] < a[j+1]:
                break
            a[j], a[j+1] = a[j+1], a[j]    
        l = i

    res.append(' '.join(map(str, a)))

    print(*res)