Problem: Phoenix loves beautiful arrays. An array is beautiful if all its subarrays of length k have the same sum. A subarray of an array is any sequence of consecutive elements.

Phoenix currently has an array a of length n. He wants to insert some number of integers, possibly zero, into his array such that it becomes beautiful. The inserted integers must be between 1 and n inclusive. Integers may be inserted anywhere (even before the first or after the last element), and he is not trying to minimize the number of inserted integers.

Input

The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 50) — the number of test cases.

The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 100).

The second line of each test case contains n space-separated integers (1 ≤ a_i ≤ n) — the array that Phoenix currently has. This array may or may not be already beautiful.

Output

For each test case, if it is impossible to create a beautiful array, print -1. Otherwise, print two lines.

The first line should contain the length of the beautiful array m (n ≤ m ≤ 10^4). You don't need to minimize m.

The second line should contain m space-separated integers (1 ≤ b_i ≤ n) — a beautiful array that Phoenix can obtain after inserting some, possibly zero, integers into his array a. You may print integers that weren't originally in array a.

If there are multiple solutions, print any. It's guaranteed that if we can make array a beautiful, we can always make it with resulting length no more than 10^4.

Example

Input


4
4 2
1 2 2 1
4 3
1 2 2 1
3 2
1 2 3
4 4
4 3 4 2


Output


5
1 2 1 2 1
4
1 2 2 1
-1
7
4 3 2 1 4 3 2

Note

In the first test case, we can make array a beautiful by inserting the integer 1 at index 3 (in between the two existing 2s). Now, all subarrays of length k=2 have the same sum 3. There exists many other possible solutions, for example: 

  * 2, 1, 2, 1, 2, 1 
  * 1, 2, 1, 2, 1, 2 



In the second test case, the array is already beautiful: all subarrays of length k=3 have the same sum 5.

In the third test case, it can be shown that we cannot insert numbers to make array a beautiful.

In the fourth test case, the array b shown is beautiful and all subarrays of length k=4 have the same sum 10. There exist other solutions also.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.*;


 

  public class C {
	
	InputStream is;
	static PrintWriter out;
	
	 
    void solve()
    {
		int t=ni();
		while(t-->0)
		{
			int n=ni(),k=ni();
			int a[]=na(n);
			int freq[]=new int[10001];
			int cum[]=new int[n];
			cum[0]=a[0];
			int ff=0;
			int mx=-100000000;
			HashSet<Integer> hs=new HashSet<Integer>();
			int ctr=0;
			
			for(int i=0;i<n;i++)
			{
				freq[a[i]]++;
				if(hs.contains(a[i])==false){
				hs.add(a[i]);
				ctr++;
				}
				mx=Math.max(mx,freq[a[i]]);
				if(a[i]>k)
					ff=-1;
			}
			/*for(int i=1;i<n;i++)
			{
				cum[i]=cum[i-1]+a[i];
				
			}
			
			int sm=(n)*(n+1);
			sm/=2;
			int ff=0;
			pa(cum);
			
				for(int j=0;j<n-k+1;j++)
				{
					if(j==0)
					{
						if(cum[j+k-1]>=sm)
						{
							ff=1;
							break;
						}
					}
					else
					{
						if((cum[j+k-1]-cum[j-1])>=sm)
						{
							ff=1;
							break;
						}
					}
				}
			*/
			
			if(ctr>k)
			{
				out.println(-1);
				continue;
			}
			out.println(n*k);
			
			
			int it=k-1;
			int ans[]=new int[k];
			
			for(int val:hs)
			{
				ans[it--]=val;
			}
			
			while(it>=0)
			{
				ans[it]=1;
				it--;
			}
			
		for(int i=0;i<n;i++)
			pa(ans);
			
			out.println();
		}
		
	}
		
	
	
	
	
	
	
	
 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	
	
 
 
	static class pair
	{
		int k;
		long v;
		pair(int kk,long vv)
		{
			k=kk;;
			v=vv;
		}
	}
	  
	  
	
    //---------- I/O Template ----------
    
    public static void main(String[] args) { new C().run(); }
    void run() { 
        is = System.in; 
        out = new PrintWriter(System.out);
        solve();
        out.flush();
    }
    
    byte input[] = new byte[1024];
    int len = 0, ptr = 0;
    
    int readByte() { 
        if(ptr >= len) { ptr = 0; 
            try { len = is.read(input); } 
            catch(IOException e) { throw new InputMismatchException(); } 
            if(len <= 0) { return -1; } 
        } return input[ptr++];
    }
    boolean isSpaceChar(int c) { return !( c >= 33 && c <= 126 ); }
    int skip() { 
        int b = readByte(); 
        while(b != -1 && isSpaceChar(b)) { b = readByte(); } 
        return b;
    }
    
    char nc() { return (char)skip(); }
    String ns() { 
        int b = skip(); 
        StringBuilder sb = new StringBuilder(); 
        while(!isSpaceChar(b)) { sb.appendCodePoint(b); b = readByte(); } 
        return sb.toString();
    }
    String nLine() { 
        int b = skip(); 
        StringBuilder sb = new StringBuilder(); 
        while( !(isSpaceChar(b) && b != ' ') ) { sb.appendCodePoint(b); b = readByte(); } 
        return sb.toString();
    }
    int ni() { 
        int n = 0, b = readByte(); 
        boolean minus = false; 
        while(b != -1 && !( (b >= '0' && b <= '9') || b == '-')) { b = readByte(); } 
        if(b == '-') { minus = true; b = readByte(); } 
        if(b == -1) { return -1; }  //no input 
        while(b >= '0' && b <= '9') { n = n * 10 + (b - '0'); b = readByte(); } 
        return minus ? -n : n;
    }
    long nl() { 
        long n = 0L;    int b = readByte(); 
        boolean minus = false; 
        while(b != -1 && !( (b >= '0' && b <= '9') || b == '-')) { b = readByte(); } 
        if(b == '-') { minus = true; b = readByte(); } 
        while(b >= '0' && b <= '9') { n = n * 10 + (b - '0'); b = readByte(); } 
        return minus ? -n : n;
    }
 
    double nd() { return Double.parseDouble(ns()); }
    float nf() { return Float.parseFloat(ns()); }
    int[] na(int n) { 
        int a[] = new int[n]; 
        for(int i = 0; i < n; i++) { a[i] = ni(); } 
        return a;
    }
    char[] ns(int n) { 
        char c[] = new char[n]; 
        int i, b = skip(); 
        for(i = 0; i < n; i++) { 
            if(isSpaceChar(b)) { break; } 
            c[i] = (char)b; b = readByte(); 
        } return i == n ? c : Arrays.copyOf(c,i);
    }
	static void pa(int a[])
	{
		for(int i=0;i<a.length;i++)
		out.print(a[i]+" ");
	
	}
	 static void pa(long a[])
	{
		for(int i=0;i<a.length;i++)
		out.print(a[i]+" ");
		out.println();
	}
	
}