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:
//2021 719

// jzzhao

import java.util.*;
import java.io.*;

public class Solution {
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter pw = new PrintWriter(System.out);
       
        int tc = sc.ni();
        for (int rep = 0; rep < tc; rep++) {
            //pw.println(solve(sc,pw));
            solve(sc,pw);
        }
        
        
        pw.close();
    }
    /*

    FS Methods:

    - next(): next element in string
    - nextLine(): nextline in string
    - ni(): next Integer
    - nd(): next Double
    - nl(): next Long
    - intArray(n): 
    - longArray(n): 
    - to2di(m,n):
    - to2dl(m,n):



    */
   

          
    public static void solve(FastScanner sc, PrintWriter pw) {

        int n = sc.ni();
        int[] arr = sc.intArray(n);
        boolean[] check = new boolean[n+1];

        int cur = 1;
        int l = 0;
        int r = 0;

        while(r<n){
            while(r<n&&arr[r]!=cur){
                check[arr[r]] = true;
                r++;
            }
            if(r==n){
                break;
            }
            if(r==l){
                r++;
                l++;
                cur++;
                continue;
            }
            //pw.println(l+" "+r+" : "+cur);
            // swap
            int tmp = r-1;
            while(tmp>=l&&arr[tmp]>cur){
                arr[tmp+1] = arr[tmp];
                tmp--;
            }
            
            tmp++;
            
            arr[tmp] = cur;
            l = r;
            
            cur++;
            while(cur<=n&&check[cur]){
                cur++;
            }
            //pw.println(Arrays.toString(arr));

        }

        printArr(pw,arr);






    }

    /*
        - The following are helping method so pls do not do anything to them.
    
    */

    // public static int[][] to2d(Scanner scanner, int m, int n){
    //     int[][] ans = new int[m][n];
    //     for(int i = 0;i<m;i++){
    //         String[] r = scanner.nextLine().split("[ ]");
    //         for(int j = 0;j<n;j++){
    //             ans[i][j] = stoi(r[j]);
    //         }
    //     }
    //     return ans;
    // }

    // public static int[] toArray(Scanner scanner, int m){
    //     int[] ans = new int[m];
    //     String[] r = scanner.nextLine().split("[ ]");
    //     for(int i = 0;i<m;i++){
    //         ans[i] = stoi(r[i]);
    //     }
        
    //     return ans;
    // }

    public static void printArr(PrintWriter pw,int[] a){
        for(int i = 0;i<a.length;i++){
            pw.print(a[i]);
            if(i!=a.length-1){
                pw.print(" ");
            }
        }
        pw.println();
    }

    public static void print2d(PrintWriter pw,int[][] a){
        for(int j=0;j<a.length;j++){
            for(int i = 0;i<a[j].length;i++){
                pw.print(a[j][i]);
                if(i!=a[j].length-1){
                    pw.print(" ");
                }
            }
            pw.println(" ");
        }
        pw.println();
    }

    static int gcd(int a, int b)
    {
        if (a == 0)
            return b; 
        return gcd(b % a, a); 
    }
    static long gcd(long a, long b) {
        if (a == 0) return b;
        return gcd(b % a, a);
    }

    public static int stoi(String s){
        return Integer.parseInt(s);
    }
}

class FastScanner {
    BufferedReader br;
    StringTokenizer st;
 
    public FastScanner() {
        br = new BufferedReader(new InputStreamReader(System.in), 32768);
        st = null;
    }
 
    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
                
            }
        }
        return st.nextToken();
    }
 
    int ni() {
        return Integer.parseInt(next());
    }
 
    int[] intArray(int N) {
        int[] ret = new int[N];
        for (int i = 0; i < N; i++)
            ret[i] = ni();
        return ret;
    }

    int[][] to2di(int m, int n){
        int[][] ans = new int[m][n];
        for(int i = 0;i<m;i++){
            String[] r = nextLine().split("[ ]");
            for(int j = 0;j<n;j++){
                ans[i][j] = Integer.parseInt(r[j]);
            }
        }
        return ans;
    }

    long[][] to2dl(int m, int n){
        long[][] ans = new long[m][n];
        for(int i = 0;i<m;i++){
            String[] r = nextLine().split("[ ]");
            for(int j = 0;j<n;j++){
                ans[i][j] = Long.parseLong(r[j]);
            }
        }
        return ans;
    }
 
    long nl() {
        return Long.parseLong(next());
    }
 
    long[] longArray(int N) {
        long[] ret = new long[N];
        for (int i = 0; i < N; i++)
            ret[i] = nl();
        return ret;
    }
 
    double nd() {
        return Double.parseDouble(next());
    }
 
    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}