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 javax.swing.plaf.synth.SynthTextAreaUI;
import java.io.*;
import java.lang.reflect.Array;
import java.text.DecimalFormat;
import java.util.*;


public class Newbie {

    static InputReader sc = new InputReader(System.in);
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) throws IOException {
        solver s = new solver();
        long t = sc.ni();
        while (t > 0) {
            s.sol();
            t--;
        }
        out.close();
    }

           /* static class descend implements Comparator<pair1> {
                public int compare(pair1 o1, pair1 o2) {
                    if (o1.pop != o2.pop)
                        return (int) (o1.pop - o2.pop);
                    else
                        return o1.in - o2.in;
                }
            }*/

    static class InputReader {
        public BufferedReader br;
        public StringTokenizer token;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream), 32768);
            token = null;
        }

        public String sn() {
            while (token == null || !token.hasMoreTokens()) {
                try {
                    token = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return token.nextToken();
        }

        public int ni() {
            return Integer.parseInt(sn());
        }

        public String snl() throws IOException {
            return br.readLine();
        }

        public long nlo() {
            return Long.parseLong(sn());
        }

        public double nd() {
            return Double.parseDouble(sn());
        }

        public int[] na(int n) {
            int a[] = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = sc.ni();
            return a;
        }

        public long[] nal(int n) {
            long a[] = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = sc.nlo();
            return a;
        }
    }

    static class ascend implements Comparator<pair> {
        public int compare(pair o1, pair o2) {
            return o2.b - o1.b;
        }
    }

    static class extra {

        static boolean v[] = new boolean[100001];
        static List<Integer> l = new ArrayList<>();
        static int t;

        static void shuffle(int a[]) {
            for (int i = 0; i < a.length; i++) {
                int t = (int) Math.random() * a.length;
                int x = a[t];
                a[t] = a[i];
                a[i] = x;
            }
        }

        static void shufflel(long a[]) {
            for (int i = 0; i < a.length; i++) {
                int t = (int) Math.random() * a.length;
                long x = a[t];
                a[t] = a[i];
                a[i] = x;
            }
        }

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

        static boolean valid(int i, int j, int r, int c) {
            if (i >= 0 && i < r && j >= 0 && j < c) {
                // System.out.println(i+" /// "+j);
                return true;
            } else {
                //  System.out.println(i+" //f "+j);
                return false;
            }
        }

        static void seive() {
            for (int i = 2; i < 101; i++) {
                if (!v[i]) {
                    t++;
                    l.add(i);
                    for (int j = 2 * i; j < 101; j += i)
                        v[j] = true;
                }
            }
        }

        static int binary(LinkedList<Integer> a, long val, int n) {
            int mid = 0, l = 0, r = n - 1, ans = 0;
            while (l <= r) {
                mid = (l + r) >> 1;
                if (a.get(mid) == val) {
                    r = mid - 1;
                    ans = mid;
                } else if (a.get(mid) > val)
                    r = mid - 1;
                else {
                    l = mid + 1;
                    ans = l;
                }
            }
            return (ans + 1);
        }

        static long fastexpo(long x, long y) {
            long res = 1;
            while (y > 0) {
                if ((y & 1) == 1) {
                    res *= x;
                }
                y = y >> 1;
                x = x * x;
            }
            return res;
        }

        static long lfastexpo(long x, long y, long p) {
            long res = 1;
            x = x % p;
            while (y > 0) {
                if ((y & 1) == 1) {
                    res = (res * x) % p;
                }
                y = y >> 1;
                x = (x * x) % p;
            }
            return res;
        }

               /* void dijsktra(int s, List<pair> l[], int n) {
                    PriorityQueue<pair> pq = new PriorityQueue<>(new ascend());
                    int dist[] = new int[100005];
                    boolean v[] = new boolean[100005];
                    for (int i = 1; i <= n; i++)
                        dist[i] = 1000000000;
                    dist[s] = 0;
                    for (int i = 1; i < n; i++) {
                        if (i == s)
                            pq.add(new pair(s, 0));
                        else
                            pq.add(new pair(i, 1000000000));
                    }
                    while (!pq.isEmpty()) {
                        pair node = pq.remove();
                        v[node.a] = true;
                        for (int i = 0; i < l[node.a].size(); i++) {
                            int v1 = l[node.a].get(i).a;
                            int w = l[node.a].get(i).b;
                            if (v[v1])
                                continue;
                            if ((dist[node.a] + w) < dist[v1]) {
                                dist[v1] = dist[node.a] + w;
                                pq.add(new pair(v1, dist[v1]));
                            }
                        }
                    }
                }*/
    }

    static class pair {
        int a;
        int b;

        public pair(int a, int b) {
            this.a = a;
            this.b = b;
        }
    }

    static class pair1 {
        pair p;
        int in;

        public pair1(pair a, int n) {
            this.p = a;
            this.in = n;
        }
    }

    static int inf = 5000013;

    static class solver {
        DecimalFormat df = new DecimalFormat("0.00000000");
        extra e = new extra();
        long mod = (long) (1000000007);
        int g[][];

        void sol() throws IOException {
            int n = sc.ni();
            int a[] = new int[n];
            HashMap<Integer, Integer> map = new HashMap<>();
            for (int i = 0; i < n; i++) {
                a[i] = sc.ni();
                map.put(a[i], i);
            }
            int b[] = new int[n];
            for (int i = 1; i <= n; i++) {
                int j=map.get(i);
                while(j>0 && a[j]<a[j-1] && b[j-1]!=-1)
                {
                    map.put(a[j],j-1);
                    map.put(a[j-1],j);
                    int t=a[j];
                    a[j]=a[j-1];
                    a[j-1]=t;
                    b[j-1]=-1;
                    j--;
                }
            }
            for (int i = 0; i < n; i++)
                System.out.print(a[i] + " ");
            System.out.println();
        }
    }
}