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.io.*;
import java.util.*;
 
 
 
public class Main {
	
	static InputReader in = new InputReader(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	
	static int oo = (int)1e9;
	static int mod = 1_000_000_007;
	
	static int[] di = {1, 0, 0, -1};
	static int[] dj = {0, -1, 1, 0};
	
	static int M = 200005;
	static double EPS = 1e-5;
	
	public static void main(String[] args) throws IOException {

		int t = in.nextInt();
		while(t --> 0) {
			int n = in.nextInt();
			int[] a = in.nextIntArray(n);
			for(int i = 0; i < n; ) {
				int min = n + 1;
				int minI = -1;
				for(int j = i; j < n; ++j) {
					if(a[j] < min) {
						min = a[j];
						minI = j;
					}
				}
				for(int j = minI; j > i; --j) {
					int tmp = a[j];
					a[j] = a[j-1];
					a[j-1] = tmp;
				}
				i = i == minI ? i+1 : minI;
			}
			for(int x : a)
				System.out.print(x + " ");
			System.out.println();
		}
		
		out.close();
	}
	
	static boolean inside(int i, int j, int n, int m) {
		return i >= 0 && i < n && j >= 0 && j < m;
	}
	
	
	static long pow(long a, long n, long mod) {
		if(n == 0)
			return 1;
		if(n % 2 == 1)
			return a * pow(a, n-1, mod) % mod;
		long x = pow(a, n / 2, mod);
		return x * x % mod;
	}
	
	
	static class SegmentTree {
		int n;
		int[] a;
		int[] seg;
		int DEFAULT_VALUE = 0;
		
		public SegmentTree(int[] a, int n) {
			super();
			this.a = a;
			this.n = n;
			seg = new int[n * 4 + 1];
			build(1, 0, n-1);
		}
		
		private int build(int node, int i, int j) {
			if(i == j) {
				return seg[node] = a[i];
			}
			int first = build(node * 2, i, (i+j) / 2);
			int second = build(node * 2 + 1, (i+j) / 2 + 1, j);
			return seg[node] = combine(first, second);
		}
		
		int update(int k, char value) {
			return update(1, 0, n-1, k, value);
		}
		
		private int update(int node, int i, int j, int k, char value) {
			if(k < i || k > j)
				return seg[node];
			if(i == j && j == k) {
				a[k] = value;
				return seg[node] = value;
			}
			int m = (i + j) / 2;
			int first = update(node * 2, i, m, k, value);
			int second = update(node * 2 + 1, m + 1, j, k, value);
			return seg[node] = combine(first, second);
		}
		
		int query(int l, int r) {
			return query(1, 0, n-1, l, r);
		}
		
		private int query(int node, int i, int j, int l, int r) {
			if(l <= i && j <= r)
				return seg[node];
			if(j < l || i > r)
				return DEFAULT_VALUE;
			int m = (i + j) / 2;
			int first = query(node * 2, i, m, l, r);
			int second = query(node * 2 + 1, m+1, j, l, r);
			return combine(first, second);
		}
 
		private int combine(int a, int b) {
			return Math.max(a, b);
		}
	}
	
	static class DisjointSet {
		int n;
		int[] g;
		int[] h;
		public DisjointSet(int n) {
			super();
			this.n = n;
			g = new int[n];
			h = new int[n];
			for(int i = 0; i < n; ++i) {
				g[i] = i;
				h[i] = 1;
			}
		}
		int find(int x) {
			if(g[x] == x)
				return x;
			return g[x] = find(g[x]);
		}
		void union(int x, int y) {
			x = find(x); y = find(y);
			if(x == y)
				return;
			if(h[x] >= h[y]) {
				g[y] = x;
				if(h[x] == h[y])
					h[x]++;
			}
			else {
				g[x] = y;
			}
		}
	}
	
	
	static int[] getPi(char[] a) {
		int m = a.length;
		int j = 0;
		int[] pi = new int[m];
		for(int i = 1; i < m; ++i) {
			while(j > 0 && a[i] != a[j])
				j = pi[j-1];
			if(a[i] == a[j]) {
				pi[i] = j + 1;
				j++;
			}
		}
		return pi;
	}
	
	static long lcm(long a, long b) {
		return a * b / gcd(a, b);
	}
	
	static boolean nextPermutation(int[] a) {
		for(int i = a.length - 2; i >= 0; --i) {
			if(a[i] < a[i+1]) {
				for(int j = a.length - 1; ; --j) {
					if(a[i] < a[j]) {
						int t = a[i];
						a[i] = a[j];
						a[j] = t;
						for(i++, j = a.length - 1; i < j; ++i, --j) {
							t = a[i];
							a[i] = a[j];
							a[j] = t;
						}
						return true;
					}
				}
			}
		}
		return false;
	}
	
	static void shuffle(Object[] a) {
		Random r = new Random();
		for(int i = a.length - 1; i > 0; --i) {
			int si = r.nextInt(i);
			Object t = a[si];
			a[si] = a[i];
			a[i] = t;
		}
	}
	
	static void shuffle(int[] a) {
		Random r = new Random();
		for(int i = a.length - 1; i > 0; --i) {
			int si = r.nextInt(i);
			int t = a[si];
			a[si] = a[i];
			a[i] = t;
		}
	}
	
	static void shuffle(long[] a) {
		Random r = new Random();
		for(int i = a.length - 1; i > 0; --i) {
			int si = r.nextInt(i);
			long t = a[si];
			a[si] = a[i];
			a[i] = t;
		}
	}
	
	static int lower_bound(int[] a, int n, int k) {
		int s = 0;
		int e = n;
		int m;
		while (e - s > 0) {
			m = (s + e) / 2;
			if (a[m] < k)
				s = m + 1;
			else
				e = m;
		}
		return e;
	}
	static int lower_bound(long[] a, int n, long k) {
		int s = 0;
		int e = n;
		int m;
		while (e - s > 0) {
			m = (s + e) / 2;
			if (a[m] < k)
				s = m + 1;
			else
				e = m;
		}
		return e;
	}
	
	static int gcd(int a, int b) {
		return b == 0 ? a : gcd(b, a % b);
	}
	static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}
	
	static class Pair implements Comparable<Pair> {
		int first, second;
 
		public Pair(int first, int second) {
			super();
			this.first = first;
			this.second = second;
		}
 
		@Override
		public int compareTo(Pair o) {
			return this.first != o.first ? this.first - o.first : this.second - o.second;
		}
		
//		@Override
//		public int compareTo(Pair o) {
//			return this.first != o.first ? o.first - this.first : o.second - this.second;
//		}
 
		@Override
		public int hashCode() {
			final int prime = 31;
			int result = 1;
			result = prime * result + first;
			result = prime * result + second;
			return result;
		}
 
		@Override
		public boolean equals(Object obj) {
			if (this == obj)
				return true;
			if (obj == null)
				return false;
			if (getClass() != obj.getClass())
				return false;
			Pair other = (Pair) obj;
			if (first != other.first)
				return false;
			if (second != other.second)
				return false;
			return true;
		}
	}
}
 
 
 
class InputReader {
 
	private final InputStream stream;
	private final byte[] buf = new byte[8192];
	private int curChar, snumChars;
 
	public InputReader(InputStream st) {
		this.stream = st;
	}
 
	public int read() {
		if (snumChars == -1)
			throw new InputMismatchException();
		if (curChar >= snumChars) {
			curChar = 0;
			try {
				snumChars = stream.read(buf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (snumChars <= 0)
				return -1;
		}
		return buf[curChar++];
	}
 
	public int nextInt() {
		int c = read();
		while (isSpaceChar(c)) {
			c = read();
		}
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		int res = 0;
		do {
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}
 
	public long nextLong() {
		int c = read();
		while (isSpaceChar(c)) {
			c = read();
		}
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		long res = 0;
		do {
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}
 
	public int[] nextIntArray(int n) {
		int a[] = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = nextInt();
		}
		return a;
	}
 
	public String readString() {
		int c = read();
		while (isSpaceChar(c)) {
			c = read();
		}
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isSpaceChar(c));
		return res.toString();
	}
 
	public String nextLine() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isEndOfLine(c));
		return res.toString();
	}
 
	public boolean isSpaceChar(int c) {
		return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	}
 
	private boolean isEndOfLine(int c) {
		return c == '\n' || c == '\r' || c == -1;
	}
 
}