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.*;

import static java.lang.Math.*;

public class Solution2 {

    //------------------------------------------CONSTANTS---------------------------------------------------------------
    public static final int MOD = (int) (1e9 + 7);

    //---------------------------------------------I/0------------------------------------------------------------------

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(
                    new InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() { return Integer.parseInt(next()); }

        long nextLong() { return Long.parseLong(next()); }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    public static void print(String s) {
        System.out.println(s);
    }

    public static void main(String[] args) {
        FastReader sc = new FastReader();

        int t = 1;
        t = sc.nextInt();

        for(int i = 0; i<t; i++) {
            System.out.println(solve(sc));
        }
    }

    //------------------------------------------------SOLVE-------------------------------------------------------------

    public static String solve(FastReader sc) {
        int n = sc.nextInt();
        List<Integer> a = new ArrayList<>();

        for (int i = 0; i<n; i++) {
            a.add(sc.nextInt());
        }

        int sum = 0;

        int ind = 0;
        for (int i = 1; i<=n; i++) {
            int firstIdx = a.indexOf(i);
            if (firstIdx <= ind) {
                if (firstIdx == ind) {
                    ind++;
                }
                continue;
            }
            int idx = firstIdx;
            while (idx > ind && sum < n-1) {
                swap(a, idx, idx-1);
                idx--;
                sum++;
            }
            if (sum >= n-1) {
                break;
            }
            ind = firstIdx;
        }

        StringBuilder res = new StringBuilder("");
        for (Integer c: a) {
            res.append(c + " ");
        }
        return res.toString();
    }

    public static void swap(List<Integer> a, int i, int j) {
        int temp = a.get(i);
        a.set(i, a.get(j));
        a.set(j, temp);
    }

    //-----------------------------------------------FUNCTIONS----------------------------------------------------------

    public static void printArray(int[] a) {
        for(int i = 0; i<a.length; i++) {
            System.out.print(a[i] + " ");
        }
    }

    // first element that is not less than target
    public static int lowerBound(int[] arr, int begin, int end, int target) {
        while(begin < end) {
            int mid = begin + (end - begin) / 2;
            // When the element is less than the mid target
            if(arr[mid] < target)
                // begin to mid + 1, arr [begin] value is less than or equal target
                begin = mid + 1;
                // When the mid target element when greater than or equal
            else if(arr[mid] >= target)
                end = mid;
        }
        return begin;
    }

    // first element that is greater than target
    public static int upperBound(int[] arr, int begin, int end, int target) {
        while(begin < end) {
            int mid = begin + (end - begin) / 2;
            if(arr[mid] <= target)
                begin = mid + 1;
            else if(arr[mid] < target)
                end = mid;
        }
        return begin;
    }

    // generate permutations (call search and access through permutations)
    public static class Permutations {
        List<List<Integer>> permutations = new ArrayList<>();
        List<Integer> permutation = new ArrayList<>();
        boolean[] taken;
        int[] a;

        public Permutations(int[] a) {
            this.a = a;
            taken = new boolean[a.length];
        }

        void search() {
            if (permutation.size()==a.length) {
                permutations.add(new ArrayList<>(permutation));
            } else {
                for (int i = 0; i<a.length; i++) {
                    if (taken[i]) {
                        continue;
                    }
                    taken[i] = true;
                    permutation.add(a[i]);
                    search();
                    taken[i] = false;
                    permutation.remove(permutation.size()-1);
                }
            }
        }
    }

    // generate subsets (call search and access subsets)
    public static class Subsets {
        List<List<Integer>> subsets = new ArrayList<>();
        List<Integer> subset = new ArrayList<>();

        void search(int[] a, int x) {
            if (x == a.length) {
                subsets.add(new ArrayList<>(subset));
            } else {
                subset.add(a[x]);
                search(a, x+1);
                subset.remove(subset.size() - 1);
                search(a, x+1);
            }
        }
    }

    public static int[] prefixSum(int[] a) {
        int n = a.length;
        int[] ps = new int[n];
        ps[0] = a[0];

        for (int i = 1; i<n; i++) {
            ps[i] = ps[i-1] + a[i];
        }

        return ps;
    }

    public static int[][] prefixSum2D(int[][] a) {
        int n = a.length;
        int m = a[0].length;
        int[][] ps = new int[n][m];
        ps[0][0] = a[0][0];

        for (int j = 1; j<m; j++) {
            ps[0][j] = ps[0][j-1] + a[0][j];
        }
        for (int i = 1; i<n; i++) {
            ps[i][0] = ps[i-1][0] + a[i][0];
        }

        for (int i = 1; i<n; i++) {
            for (int j = 1; j<m; j++) {
                ps[i][j] = ps[i-1][j] + ps[i][j-1] - ps[i-1][j-1] + a[i][j];
            }
        }

        return ps;
    }

    public static double log2(double a) {
        return log(a)/log(2);
    }

    //----------------------------------------DATA-STRUCTURES-----------------------------------------------------------

    // Range queries (i.e. RMQ) in O(1), others with functions in O(log n)
    public static class SparseTable {
        int[][] st;
        int n;
        int k;

        public SparseTable(int[] a) {
            n = a.length;
            k = (int) log2(n);
            st = new int[n][k+1];

            for (int i = 0; i<n; i++) {
                st[i][0] = a[i];
            }

            // 1<<j == 2^j
            for (int j = 1; j<=k; j++) {
                for (int i = 0; i + (1 << j) <= n; i++) {
                    st[i][j] = min(st[i][j-1], st[i + (1 << (j - 1))][j - 1]);
                }
            }
        }

        //not always available O(log n) -- change the constructor to use it
        public long sumQuery(int l, int r) {
            long sum = 0;
            for (int j = k; j>=0; j--) {
                if ((1 << j) <= r - l + 1) {
                    sum += st[l][j];
                    l += 1 << j;
                }
            }

            return sum;
        }

        //always available O(1) -- change this and the constructor for rangeMaxQuery
        public int rangeMinQuery(int l, int r) {
            int j = (int) log2(r - l + 1);
            int min = min(st[l][j], st[r - (1 << j) + 1][j]);
            return min;
        }

    }

    public static class BinaryIndexedTree {

        int[] bit;

        public BinaryIndexedTree(int[] a) {
            int n = a.length;
            bit = new int[n+1];
            // Store the actual values in BITree[]
            // using update()
            for(int i = 0; i < n; i++)
                updateBit(i, a[i]);
        }

        // O(log n) update delta at index in the array (do it on the array and then call this function)
        public void updateBit(int index, int delta) {
            index += 1;

            while (index <= bit.length) {
                bit[index] += delta;

                index += index & (-index);
            }
        }

        // Computes sum a[0...index] in O(log n)
        public int getSum(int index) {
            int sum = 0;

            index += 1;

            while (index > 0) {
                sum += bit[index];

                index -= index & (-index);
            }

            return sum;
        }
    }
}