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:

/**
 * Date: 04 Nov, 2019
 * Link: https://codeforces.com/contest/1256/problem/B
 * 
 * @author Prasad Chaudhari
 * @linkedIn: https://www.linkedin.com/in/prasad-chaudhari-841655a6/
 * @git: https://github.com/Prasad-Chaudhari
 */
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Arrays;

public class B_1256 {

    public static void main(String[] args) throws IOException {
        // TODO code application logic here
        FastIO in = new FastIO();
        int q = in.ni();
        while (q-- > 0) {
            int n = in.ni();
            int a[] = in.gia(n);
            int count = 0;
            boolean b[] = new boolean[n];
            for (int i = 1; i <= n; i++) {
                int k = 0;
                for (int j = 0; j < n; j++) {
                    if (a[j] == i) {
                        k = j;
                        break;
                    }
                }
                for (int j = k - 1; j >= i - 1; j--) {
                    if (!b[j] && count < n - 1 && a[j] > a[j + 1]) {
                        a[j] ^= a[j + 1];
                        a[j + 1] ^= a[j];
                        a[j] ^= a[j + 1];
                        b[j] = true;
                        count++;
                    } else {
                        break;
                    }
                }
            }
            for (int i = 0; i < n; i++) {
                in.print(a[i] + " ");
            }
            in.println("");
        }
        in.bw.flush();
    }

    static class Data implements Comparable<Data> {

        int a, b;

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

        @Override
        public int compareTo(Data o) {
            if (a == o.a) {
                return Integer.compare(b, o.b);
            }
            return Integer.compare(a, o.a);
        }

        public static void sort(int a[]) {
            Data d[] = new Data[a.length];
            for (int i = 0; i < a.length; i++) {
                d[i] = new Data(a[i], 0);
            }
            Arrays.sort(d);
            for (int i = 0; i < a.length; i++) {
                a[i] = d[i].a;
            }
        }
    }

    static class FastIO {

        private final BufferedReader br;
        private final BufferedWriter bw;
        private String s[];
        private int index;

        public FastIO() throws IOException {
            br = new BufferedReader(new InputStreamReader(System.in));
            bw = new BufferedWriter(new OutputStreamWriter(System.out, "UTF-8"));
            s = br.readLine().split(" ");
            index = 0;
        }

        public int ni() throws IOException {
            return Integer.parseInt(nextToken());
        }

        public double nd() throws IOException {
            return Double.parseDouble(nextToken());
        }

        public long nl() throws IOException {
            return Long.parseLong(nextToken());
        }

        public String next() throws IOException {
            return nextToken();
        }

        public String nli() throws IOException {
            try {
                return br.readLine();
            } catch (IOException ex) {

            }
            return null;
        }

        public int[] gia(int n) throws IOException {
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public int[] gia(int n, int start, int end) throws IOException {
            validate(n, start, end);
            int a[] = new int[n];
            for (int i = start; i < end; i++) {
                a[i] = ni();
            }
            return a;
        }

        public double[] gda(int n) throws IOException {
            double a[] = new double[n];
            for (int i = 0; i < n; i++) {
                a[i] = nd();
            }
            return a;
        }

        public double[] gda(int n, int start, int end) throws IOException {
            validate(n, start, end);
            double a[] = new double[n];
            for (int i = start; i < end; i++) {
                a[i] = nd();
            }
            return a;
        }

        public long[] gla(int n) throws IOException {
            long a[] = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nl();
            }
            return a;
        }

        public long[] gla(int n, int start, int end) throws IOException {
            validate(n, start, end);
            long a[] = new long[n];
            for (int i = start; i < end; i++) {
                a[i] = nl();
            }
            return a;
        }

        public int[][][] gwtree(int n) throws IOException {
            int m = n - 1;
            int adja[][] = new int[n + 1][];
            int weight[][] = new int[n + 1][];
            int from[] = new int[m];
            int to[] = new int[m];
            int count[] = new int[n + 1];
            int cost[] = new int[n + 1];
            for (int i = 0; i < m; i++) {
                from[i] = i + 1;
                to[i] = ni();
                cost[i] = ni();
                count[from[i]]++;
                count[to[i]]++;
            }
            for (int i = 0; i <= n; i++) {
                adja[i] = new int[count[i]];
                weight[i] = new int[count[i]];
            }
            for (int i = 0; i < m; i++) {
                adja[from[i]][count[from[i]] - 1] = to[i];
                adja[to[i]][count[to[i]] - 1] = from[i];
                weight[from[i]][count[from[i]] - 1] = cost[i];
                weight[to[i]][count[to[i]] - 1] = cost[i];
                count[from[i]]--;
                count[to[i]]--;
            }
            return new int[][][] { adja, weight };
        }

        public int[][][] gwg(int n, int m) throws IOException {
            int adja[][] = new int[n + 1][];
            int weight[][] = new int[n + 1][];
            int from[] = new int[m];
            int to[] = new int[m];
            int count[] = new int[n + 1];
            int cost[] = new int[n + 1];
            for (int i = 0; i < m; i++) {
                from[i] = ni();
                to[i] = ni();
                cost[i] = ni();
                count[from[i]]++;
                count[to[i]]++;
            }
            for (int i = 0; i <= n; i++) {
                adja[i] = new int[count[i]];
                weight[i] = new int[count[i]];
            }
            for (int i = 0; i < m; i++) {
                adja[from[i]][count[from[i]] - 1] = to[i];
                adja[to[i]][count[to[i]] - 1] = from[i];
                weight[from[i]][count[from[i]] - 1] = cost[i];
                weight[to[i]][count[to[i]] - 1] = cost[i];
                count[from[i]]--;
                count[to[i]]--;
            }
            return new int[][][] { adja, weight };
        }

        public int[][] gtree(int n) throws IOException {
            int adja[][] = new int[n + 1][];
            int from[] = new int[n - 1];
            int to[] = new int[n - 1];
            int count[] = new int[n + 1];
            for (int i = 0; i < n - 1; i++) {
                from[i] = i + 1;
                to[i] = ni();
                count[from[i]]++;
                count[to[i]]++;
            }
            for (int i = 0; i <= n; i++) {
                adja[i] = new int[count[i]];
            }
            for (int i = 0; i < n - 1; i++) {
                adja[from[i]][--count[from[i]]] = to[i];
                adja[to[i]][--count[to[i]]] = from[i];
            }
            return adja;
        }

        public int[][] gg(int n, int m) throws IOException {
            int adja[][] = new int[n + 1][];
            int from[] = new int[m];
            int to[] = new int[m];
            int count[] = new int[n + 1];
            for (int i = 0; i < m; i++) {
                from[i] = ni();
                to[i] = ni();
                count[from[i]]++;
                count[to[i]]++;
            }
            for (int i = 0; i <= n; i++) {
                adja[i] = new int[count[i]];
            }
            for (int i = 0; i < m; i++) {
                adja[from[i]][--count[from[i]]] = to[i];
                adja[to[i]][--count[to[i]]] = from[i];
            }
            return adja;
        }

        public void print(String s) throws IOException {
            bw.write(s);
        }

        public void println(String s) throws IOException {
            bw.write(s);
            bw.newLine();
        }

        public void print(int s) throws IOException {
            bw.write(s + "");
        }

        public void println(int s) throws IOException {
            bw.write(s + "");
            bw.newLine();
        }

        public void print(long s) throws IOException {
            bw.write(s + "");
        }

        public void println(long s) throws IOException {
            bw.write(s + "");
            bw.newLine();
        }

        public void print(double s) throws IOException {
            bw.write(s + "");
        }

        public void println(double s) throws IOException {
            bw.write(s + "");
            bw.newLine();
        }

        private String nextToken() throws IndexOutOfBoundsException, IOException {
            if (index == s.length) {
                s = br.readLine().split(" ");
                index = 0;
            }
            return s[index++];
        }

        private void validate(int n, int start, int end) {
            if (start < 0 || end >= n) {
                throw new IllegalArgumentException();
            }
        }
    }
}