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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import java.util.*;
import java.io.PrintStream;
import java.io.PrintWriter;
 
public class A {
 
	static final int mod = 1000000007;
	static final int temp = 998244353;
	static final long MOD = 1000000007;
	static final long M = (long)1e9+7;
 
	
	static class Pair implements Comparable<Pair>
	{
		int first, second;
		public Pair(int aa, int bb)
		{
			first = aa; second = bb;
		}
		public int compareTo(Pair p)
		{
//			if(a == p.a) return b - p.b;
//			return a - p.a;
			if(first == p.first) return (int)(second - p.second);
			return (int)(first - p.first);
		}
	}
	
	/* IO for List of Arrays
	 * ArrayList<Integer>[] list = new ArrayList[m];
		for(int i = 0;i<m;i++)
		{
			list[i] = new ArrayList<>();
		}
	 */
	
	/*
	 * IO FOR 2D GRID IN JAVA
	 * 
	 * 		char[][] grid = new char[n][n];
	 * 		for(int i = 0;i<n;i++) grid[i] = sc.next().toCharArray();
 
	 * */
	
	static class Reader {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		List<Integer> readList(int n)
		{
			List<Integer> l = new ArrayList<>();
			for(int i = 0;i<n;i++) l.add(nextInt());
			return l;
		}
		
		long[] longReadArray(int n) throws IOException {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	
 
	public static boolean isPrime(long n) {
		if(n == 1)
		{
			return false;
		}
		for(long i = 2;i*i<=n;i++)
		{
			if(n%i == 0)
			{
				return false;
			}
		}
		return true;
	}
	
	public static List<Integer> Sieve(int n)
	{
		boolean prime[] = new boolean[n+1];
		Arrays.fill(prime, true);
		List<Integer> l = new ArrayList<>();
		
		
		for (int p=2; p*p<=n; p++) 
		{ 
			if (prime[p] == true) 
			{ 	
			    for(int i=p*p; i<=n; i += p) 
				{
				    prime[i] = false; 
				}	
			} 
		} 
 
		for (int p=2; p<=n; p++) 
		{
		    if (prime[p] == true)
		    {
		       l.add(p); 
		    }
		}
		
		return l;
	}
	
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	} 
 
	public static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
	
	public static long LongGCD(long a, long b)
	{
		if(b == 0)
			 return a;
			
			else
			return LongGCD(b,a%b);
	}
	
	public static long LongLCM(long a, long b)
    {
        return (a / LongGCD(a, b)) * b;
    }
	
	public static int phi(int n)  //euler totient function
    { 
        int result = 1; 
        for (int i = 2; i < n; i++) 
            if (gcd(i, n) == 1) 
                result++; 
        return result; 
    }
	
	    
	public static int[] computePrefix(int arr[], int n)
	{
		int[] prefix = new int[n];
		prefix[0] = arr[0];
		for(int i = 1;i<n;i++)
		{
			prefix[i] = prefix[i-1]+arr[i];
		}
		
		return prefix;
	}
	
	
	public static long fastPow(long x, long n) //include mod at each step if asked and in args of fn too
	{
		if(n == 0)
			return 1;
		else if(n%2 == 0)
			return fastPow(x*x,n/2);
		else
			return x*fastPow(x*x,(n-1)/2);
	}
	
	
	   public static long modPow(long x, long y, long p)
	    {
	 
	        long res = 1;
	 
	        x = x % p;
	 
	        while (y > 0) {
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	 
	        return res;
	    }
	   
	   static long modInverse(long n, long p)
	    {
	        return modPow(n, p - 2, p);
	    }
	 
	    // Returns nCr % p using Fermat's little theorem.
	    
	    public static long nCr(long n, long r,long p)
	    {
	          if (n<r) 
	              return 0;
	          
	        if (r == 0)
	            return 1;
	 
	        long[] fac = new long[(int)(n) + 1];
	        fac[0] = 1;
	 
	        for (int i = 1; i <= n; i++)
	            fac[i] = fac[i - 1] * i % p;
	 
	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
	                % p * modInverse(fac[(int)(n - r)], p)
	                % p)
	            % p;
	    }
	    
	    public static long[] modInvArr(int n, long mod)
	    {
	    	long[] ans = new long[n+1];
			ans[1] = 1;
			for(int i = 2;i <= n;i++)
			{
				ans[i] = (mod - (mod/i) * ans[(int)(mod%i)])%mod;
			}
			return ans;
	    }
 
		public static void Sort(long[] a) {
			List<Long> l=new ArrayList<>();
			for (long i:a) l.add(i);
			Collections.sort(l);
			//Collections.reverse(l);  //Use to Sort decreasingly 
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		public static void ssort(char[] a) {
			List<Character> l=new ArrayList<>();
			for (char ch : a) l.add(ch);
			Collections.sort(l);
			//Collections.reverse(l);  //Use to Sort decreasingly 
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		//Modular Operations for Addition and Multiplication. 
		
		   public static long perfomMod(long x){
		        return ((x%M + M)%M);
		    }
		   public static long addMod(long a, long  b){
		        return perfomMod(perfomMod(a)+perfomMod(b));
		    }
		   public static long mulMod(long  a, long b){
		        return perfomMod(perfomMod(a)*perfomMod(b));
		   }
		
		   
	public static void main(String[] args) throws IOException 
	{
		Reader sc=new Reader();
		PrintWriter out = new PrintWriter(System.out);
		
		int t = sc.nextInt();
		while(t-- > 0)
		{
			int n = sc.nextInt();
			int[] a = sc.readArray(n);
			int[] b = new int[n];
			for(int x : a)
			{
				b[x-1] = 0;
			}
			for(int i=n-2;i>=0;i--)
			{
				for(int j = i;j<n-1;j++)
				{
					if(a[j] > a[j+1] && b[j] == 0)
					{
						int tx = a[j];
						a[j] = a[j+1];
						a[j+1] = tx;
						b[j] = 1;
					}
				}
			}
			
			for(int i = 0;i<n;i++)
			{
				System.out.print(a[i] + " ");
			}
			System.out.println();
		}
			
	}	
}