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.BigInteger;
public class cf1 {
    static long mod = (long)1e9 + 7;
    static long mod1 = 998244353;
    static FastScanner f;
    static PrintWriter pw = new PrintWriter(System.out);
    static Scanner S = new Scanner(System.in);
    static long x0; static long y0;
    static int inf = (int)(1e9);
    static long iinf = (long)(1e18);
    static void solve()throws NumberFormatException , IOException {
        int n = f.ni();
        int arr[] = new int[n + 1];
        boolean vis[] = new boolean[n + 1];
        for (int i = 1; i <= n; ++i) {
            arr[i] = f.ni();
        }
        for (int i = 1; i <= n; ++i) {
            if (arr[i] == i) continue;
            int idx = -1;
            for (int j = i + 1; j <= n; ++j) {
                if (arr[j] == i) {
                    idx = j;
                    break;
                }
            }
            //pn("i = " + i + "idx = " + idx);
            for (int j = idx - 1; j >= i; --j) {
                if (!vis[j] && arr[j] > arr[j + 1]) {
                    int x = arr[j];
                    arr[j] = arr[j + 1];
                    arr[j + 1] = x;
                    vis[j] = true;
                }
                else break;
            }
        }
        for (int i = 1; i <= n; ++i) p(arr[i] + " ");
        pn("");
    }
    public static void main(String[] args)throws NumberFormatException , IOException {
        init(); 
        int t = f.ni();
        while(t --> 0) {solve();}
        pw.flush(); 
        pw.close();  
    }
        
/******************************END OF MAIN PROGRAM*******************************************/
    public static void init()throws IOException{if(System.getProperty("ONLINE_JUDGE")==null){f=new FastScanner("");}else{f=new FastScanner(System.in);}}
    public static class FastScanner {
        BufferedReader br;StringTokenizer st;
        FastScanner(InputStream stream){try{br=new BufferedReader(new InputStreamReader(stream));}catch(Exception e){e.printStackTrace();}}
        FastScanner(String str){try{br=new BufferedReader(new FileReader("!a.txt"));}catch(Exception e){e.printStackTrace();}}
        String next(){while(st==null||!st.hasMoreTokens()){try{st=new StringTokenizer(br.readLine());}catch(IOException e){e.printStackTrace();}}return st.nextToken();}
        String nextLine()throws IOException{return br.readLine();}int ni(){return Integer.parseInt(next());}long nl(){return Long.parseLong(next());}double nd(){return Double.parseDouble(next());}
    }
    public static void pn(Object o){pw.println(o);}
    public static void p(Object o){pw.print(o);}
    public static void pni(Object o){pw.println(o);pw.flush();}
    static int gcd(int a,int b){if(b==0)return a;else{return gcd(b,a%b);}}
    static long gcd(long a,long b){if(b==0l)return a;else{return gcd(b,a%b);}}
    static long lcm(long a,long b){return (a*b/gcd(a,b));}
    static long exgcd(long a,long b){if(b==0){x0=1;y0=0;return a;}long temp=exgcd(b,a%b);long t=x0;x0=y0;y0=t-a/b*y0;return temp;}
    static long pow(long a,long b){long res=1;while(b>0){if((b&1)==1)res=res*a;b>>=1;a=a*a;}return res;}
    static long mpow(long a,long b){long res=1;while(b>0l){if((b&1)==1l)res=((res%mod)*(a%mod))%mod;b>>=1l;a=((a%mod)*(a%mod))%mod;}return res;}
    static long mul(long a , long b){return ((a%mod)*(b%mod)%mod);}
    static long adp(long a , long b){return ((a%mod)+(b%mod)%mod);}
    static int log2(int x){return (int)(Math.log(x)/Math.log(2));}
    static boolean isPrime(long n){if(n<=1)return false;if(n<=3)return true;if(n%2==0||n%3==0)return false;for(long i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
    static boolean isPrime(int n){if(n<=1)return false;if(n<=3)return true;if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
    static HashSet<Long> factors(long n){HashSet<Long> hs=new HashSet<Long>();for(long i=1;i<=(long)Math.sqrt(n);i++){if(n%i==0){hs.add(i);hs.add(n/i);}}return hs;}
    static HashSet<Integer> factors(int n){HashSet<Integer> hs=new HashSet<Integer>();for(int i=1;i<=(int)Math.sqrt(n);i++){if(n%i==0){hs.add(i);hs.add(n/i);}}return hs;}
    static HashSet<Long> pf(long n){HashSet<Long> ff=factors(n);HashSet<Long> ans=new HashSet<Long>();for(Long i:ff)if(isPrime(i))ans.add(i);return ans;}
    static HashSet<Integer> pf(int n){HashSet<Integer> ff=factors(n);HashSet<Integer> ans=new HashSet<Integer>();for(Integer i:ff)if(isPrime(i))ans.add(i);return ans;}
    static int[] inpint(int n){int arr[]=new int[n];for(int i=0;i<n;i++){arr[i]=f.ni();}return arr;}
    static long[] inplong(int n){long arr[] = new long[n];for(int i=0;i<n;i++){arr[i]=f.nl();}return arr;}
    static boolean ise(int x){return ((x&1)==0);}static boolean ise(long x){return ((x&1)==0);}
    static int gnv(char c){return Character.getNumericValue(c);}//No. of integers less than equal to i in ub
    static int log(long x){return x==1?0:(1+log(x/2));} static int log(int x){return x==1?0:(1+log(x/2));}
    static int upperbound(int a[],int i){int lo=0,hi=a.length-1,mid=0;int count=0;while(lo<=hi){mid=(lo+hi)/2;if(a[mid]<=i){count=mid+1;lo=mid+1;}else hi=mid-1;}return count;}
    static void sort(int[] a){ArrayList<Integer> l=new ArrayList<>();for(int i:a)l.add(i);Collections.sort(l);for(int i=0;i<a.length;++i)a[i]=l.get(i);}
    static void sort(long[] a){ArrayList<Long> l=new ArrayList<>();for(long i:a)l.add(i);Collections.sort(l);for(int i=0;i<a.length;++i)a[i]=l.get(i);}
    static void sort(ArrayList<Integer> a){Collections.sort(a);}//!Precompute fact in ncr()!
    static int nextPowerOf2(int n){int count=0;if(n>0&&(n&(n-1))==0)return n;while(n!=0){n>>=1;count += 1;}return 1<<count;} 
}