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 JAVA):

Solution:

	import java.util.*;
	import java.io.*;
	import java.math.*;
	public class Codeforces {
		static char c[];
		public static void main(String[] args) throws Exception{
			//BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
			Scanner input = new Scanner(System.in);
			int test = input.nextInt();
			while(test-->0) {
				int n = input.nextInt();
				ArrayList<Integer>list = new ArrayList<>();
				for(int i =0;i<n;i++) {int a = input.nextInt();list.add(a-1);}
				int last = 0;
				for(int i =0;i<n;i++) {
					int min = n;
					int index = -1;
					for(int j =i;j<n;j++)if(list.get(j)<min) {min = list.get(j);index = j;}
					if(index == i)continue;
					list.remove(index);
					list.add(i,min);
					i = index-1;
					
				}
				for(int it:list)System.out.print(it+1+" ");
				System.out.println();
			}
			
		}
		public static void swap(int a,int b,int arr[]) {
			int t = arr[b];
			arr[b] = arr[a];
			arr[a] = t;
		}
		
	}
	/*****************************************Notes**********************************************
	
	1.Never use String.charAt for integers=> only works till 9 if integer val is 28,39,555
	charAt would pick only the first digit. Better use Integer.parseInt() to convert string to Integer
	
	2. even empty for loop can be used in certain cases.eg 'C - Pinkie Pie Eats Patty-cakes'. 
	ie.the check function of binary search. extra code but learnt some approach = SecondThreads implementation
	
	3. The two list subsequence method eg:D. Binary String To Subsequences (works only in the case of two characters)
	
	
	4. when there is a range that is computed there are case where index out of bounds might occur to keep
	the ranges proper use min max trick=> leftfarthest = max(0,computed) && rightFarthest = min(n-1,computed)
	
	***************************************************
	5.Think in reverse order eg:puzzle game
	bfs can be used to generate all possible combinations
	*******************
	6.carefully see the bounds for example incase of time see if its 12 hour or 24 hour
	**********************************
	plan the strategy before even coding and check for test cases if it workkkkkkkkkk
	
	sliding window technique: first and sec can be zero and is used especially if two numbers satify 
	a condition then the entire range satisfies(mainly on sorted arrays).whenever a variable is defaulted before
	execution think if that would change the result for certain cases and alter the initialization of res variable accordingly
	
	************
	in count of same value pick pairs of elements =>count/2;
	in count of different variables pick pairs => Math.min(count[a],count[b]);
	
	number divsible by k = k-elements[i]%k
	
	
	
	**************
	*/