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:
/*Author: Satyajeet Singh, Delhi Technological University*/
    import java.io.*;
    import java.util.*;
    import java.text.*; 
    import java.lang.*;
    import java.math.*;
 
public class Main{
/*********************************************Constants******************************************/
    static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out));        
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static long mod=(long)1e9+7;
    static long mod1=998244353;
    static boolean sieve[];
    static ArrayList<Integer> primes;
    static long factorial[],invFactorial[];
    static ArrayList<Pair> graph[];
    static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
    static int pptr=0;
    static String st[];
/****************************************Solutions Begins***************************************/
    public static void main(String[] args) throws Exception{
        nl();
        int t=pi();
        while(t-->0){
            nl();
            int n=pi();
            nl();
            int input[]=new int[n];
        
            for(int i=0;i<n;i++){
                input[i]=pi();
            }
            int till=0;
            while(till<n){
                int min=Integer.MAX_VALUE;
                int idx=0;
                for(int i=till;i<n;i++){
                    if(min>input[i]){
                        min=input[i];
                        idx=i;
                    }
                }
                if(idx==till){
                    till++;
                    continue;
                }
            //  debug(till,idx,min);
                for(int i=idx;i>till;i--){
                    input[i]=input[i-1];
                }
                input[till]=min; 
                till=idx;  
            }
            for(int i:input)out.print(i+" ");
            out.println();
        }
/****************************************Solutions Ends**************************************************/
        out.flush();
        out.close();
    }
/****************************************Template Begins************************************************/
    static void nl() throws Exception{
        pptr=0;
        st=br.readLine().split(" ");
    }
    static void nls() throws Exception{
        pptr=0;
        st=br.readLine().split("");
    }
    static int pi(){
        return Integer.parseInt(st[pptr++]);
    }
    static long pl(){
        return Long.parseLong(st[pptr++]);
    }
    static double pd(){
        return Double.parseDouble(st[pptr++]);
    }
/***************************************Precision Printing**********************************************/
    static void printPrecision(double d){
        DecimalFormat ft = new DecimalFormat("0.000000000000000000000"); 
        out.print(ft.format(d));
    }
/**************************************Bit Manipulation**************************************************/
    static void printMask(long mask){
        System.out.println(Long.toBinaryString(mask));
    }
    static long countBit(long mask){
        long ans=0;
        while(mask!=0){
            if(mask%2==1){
                ans++;
            }
            mask/=2;
        }
        return ans;
    }
/******************************************Graph*********************************************************/
    static void Makegraph(int n){
        graph=new ArrayList[n];
        for(int i=0;i<n;i++){
            graph[i]=new ArrayList<>();
        }
    }
    static void addEdge(int a,int b){
        graph[a].add(new Pair(b,1));
    }
    static void addEdge(int a,int b,int c){
        graph[a].add(new Pair(b,c));
    }    
/*********************************************PAIR********************************************************/
    static class Pair implements Comparable<Pair> {
        int u;
        int v;
        int index=-1;
        public Pair(int u, int v) {
            this.u = u;
            this.v = v;
        }
        public int hashCode() {
            int hu = (int) (u ^ (u >>> 32));
            int hv = (int) (v ^ (v >>> 32));
            return 31 * hu + hv;
        }
        public boolean equals(Object o) {
            Pair other = (Pair) o;
            return u == other.u && v == other.v;
        }
        public int compareTo(Pair other) {
            if(index!=other.index)
                return Long.compare(index, other.index);
            return Long.compare(v, other.v)!=0?Long.compare(v, other.v):Long.compare(u, other.u);
        }
        public String toString() {
            return "[u=" + u + ", v=" + v + "]";
        }
    }
/******************************************Long Pair*******************************************************/
    static class Pairl implements Comparable<Pairl> {
            long u;
            long v;
            int index=-1;
            public Pairl(long u, long v) {
                this.u = u;
                this.v = v;
            }
            public int hashCode() {
                int hu = (int) (u ^ (u >>> 32));
                int hv = (int) (v ^ (v >>> 32));
                return 31 * hu + hv;
            }
            public boolean equals(Object o) {
                Pairl other = (Pairl) o;
                return u == other.u && v == other.v;
            }
            public int compareTo(Pairl other) {
                if(index!=other.index)
                    return Long.compare(index, other.index);
                return Long.compare(v, other.v)!=0?Long.compare(v, other.v):Long.compare(u, other.u);
            }
            public String toString() {
                return "[u=" + u + ", v=" + v + "]";
            }
        }
/*****************************************DEBUG***********************************************************/
    public static void debug(Object... o) {
        if(!oj)
        System.out.println(Arrays.deepToString(o));
    }
/************************************MODULAR EXPONENTIATION***********************************************/
    static long modulo(long a,long b,long c) {
        long x=1;
        long y=a%c;
        while(b > 0){
            if(b%2 == 1){
                x=(x*y)%c;
            }
            y = (y*y)%c; // squaring the base
            b /= 2;
        }
        return  x%c;
    }
/********************************************GCD**********************************************************/
    static long gcd(long x, long y)
    {
        if(x==0)
            return y;
        if(y==0)
            return x;
        long r=0, a, b;
        a = (x > y) ? x : y; // a is greater number
        b = (x < y) ? x : y; // b is smaller number
        r = b;
        while(a % b != 0)
        {
            r = a % b;
            a = b;
            b = r;
        }
        return r;
    }
/******************************************SIEVE**********************************************************/
    static void sieveMake(int n){
        sieve=new boolean[n];
        Arrays.fill(sieve,true);
        sieve[0]=false;
        sieve[1]=false;
        for(int i=2;i*i<n;i++){
            if(sieve[i]){
                for(int j=i*i;j<n;j+=i){
                    sieve[j]=false;
                }
            }
        }
        primes=new ArrayList<Integer>();
        for(int i=0;i<n;i++){
            if(sieve[i]){
                primes.add(i);
            }
        }        
    }
/********************************************End***********************************************************/
}