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:
mod=10**9+7
import sys 
sys.setrecursionlimit(10**6) 
from sys import stdin, stdout
import bisect
from bisect import bisect_left as bl              #c++ lowerbound bl(array,element)
from bisect import bisect_right as br             #c++ upperbound
import itertools
import collections
import math
import heapq
import random
def modinv(n,p):
    return pow(n,p-2,p)
def ncr(n,r,p):                        #for using this uncomment the lines calculating fact and ifact
    t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
    return t
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
def GCD(x,y):
    while(y):
        x, y = y, x % y
    return x
"""*******************************************************"""
def main():
    t=inin()
    for _ in range(t):
        n=inin()
        a=ain()
        i=0
        y=n-1
        x=0
        while(i<n):
            for k in range(x,n):
                j=a.index(k+1)
                if(j-i<=n-1 and j-i>0 and a[j]<a[i]):
                    a.pop(j)
                    a.insert(i,k+1)
                    i=j-1
                    break
            i+=1
            x+=1
        print(*a)
######## Python 2 and 3 footer by Pajenegod and c1729
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A
 
if __name__== "__main__":
  main()