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.util.*;
import java.io.*;
public class lp{  
public static void main(String[] args){

int t = ni();
for(int o=0;o<t;o++)
{
  int n = ni();
  int k = ni();
  int a[] = ai(n);
  int ar[] = new int[k];
  int is=-1,f=1;
  HashSet<Integer> h =new HashSet();
  for(int i=0;i<n;i++)
    {
         if(!h.contains(a[i])){
              is++;
              if(is==k)
               {f=0;break;}
             h.add(a[i]);
             ar[is]=a[i];
         }   
       
    }
  if(f==0)
     pn(-1);
  else{
    pn(n*k);
    is++;
    while(is<k)
       ar[is++]=1; 
    for(int i=0;i<n;i++)
      for(int j=0;j<k;j++)
        p(ar[j]+" ");
    pn("");
  }
}
  

   out.flush();
}
 
static class pair
	{
		int a,b;
		pair(){}
		pair(int c,int d){a=c;b=d;}
		@Override
		public int hashCode(){return (a+" "+b).hashCode();}
		@Override public boolean equals(Object c){if(a==((pair)c).a&&b==((pair)c).b)return true;return false;}
	}
static int get(int index)
        {
                return (index&(-index));
        }
static void add(int bit[],int index,int val)
        {
             while(index<bit.length)
               {
                    bit[index] = bit[index]+val;
                    index = index + get(index);
               }
        }
static int get(int bit[],int index)
         {
               int sum=0;
               while(index>0)
                {
                    sum = sum+bit[index];
                    index = index-get(index);
                }
              return sum;
         }
static void add(long bit[],int index,long val)
        {
             while(index<bit.length)
               {
                    bit[index] = bit[index]+val;
                    index = index + get(index);
               }
        }
static long get(long bit[],int index)
         {
               long sum=0;
               while(index>0)
                {
                    sum = sum+bit[index];
                    index = index-get(index);
                }
              return sum;
         } 
static void compress(int a[])
         {      
                int n = a.length;
                Integer ar[] = new Integer[n];
                for(int i=0;i<n;i++)
                     ar[i] = a[i];
                Arrays.sort(ar);
                HashMap<Integer,Integer> hm = new HashMap();
                int st=1;
                for(int i=0;i<n;i++)
                    if(!hm.containsKey(ar[i]))
                        hm.put(ar[i],st++);
                for(int i=0;i<n;i++)
                   a[i] = hm.get(a[i]);
         }
static int find(int x,int p[]){ if(x==p[x]) return x; return p[x]=find(p[x],p) ;}
static void union(int x,int y,int p[]){ int xr=find(x,p),yr=find(y,p); if(xr!=yr){ p[xr]=yr;} }
 
static int abs(int x){ return x>0 ? x : -x; }
static long abs(long x){ return x>0 ? x : -x; }
static long gcd(long a,long b){ if(a==0)return b; if(b==0) return a;if(b%a==0) return a;return gcd(b%a,a); }
static int gcd(int a,int b){ if(a==0)return b; if(b==0) return a; if(b%a==0) return a;return gcd(b%a,a); }
static long min(long a,long b){ return a<b ? a : b; }
static int min(int a,int b){ return a<b ? a : b; }
static long max(long a,long b){ return a>b ? a : b; }
static int max(int a,int b){ return a>b ? a : b; }
static long pow(long a,long b,long md){ long ans=1;  while(b>0){if(b%2==1) ans = (ans*a)%md; a = (a*a)%md; b = b/2;} return ans; }
static int log2n(long a){ int te=0; while(a>0) {a>>=1; ++te; } return te; }

static void reverse(int a[]){int i1=0,i2=a.length-1; while(i1<i2){int t = a[i1];a[i1]=a[i2];a[i2]=t;i1++;i2--;}  } 
static void reverse(long a[]){ int i1=0,i2=a.length-1; while(i1<i2){ long t = a[i1]; a[i1]=a[i2];a[i2]=t;i1++;i2--;}  } 

static void subtract_1(char s[]){if(s[0]=='0') return;int n = s.length,i=n-1;while(s[i]=='0')i--;s[i] =(char)((int)(s[i]-'0') + 47);for(int j=i+1;j<n;j++)s[j]='9';}
static void reverse(char s[]){int i1=0,i2=s.length-1; while(i1<i2){char t=s[i1];s[i1]=s[i2];s[i2]=t; i1++;i2--;} }
static String reverse(String r){String s = "";int i= r.length()-1; while(i>=0){ s=s+r.charAt(i);i--;}return s;}
static boolean pal(char s[]){ int i1=0,i2=s.length-1; while(i1<i2){ if(s[i1]!=s[i2]) return false; i1++;i2--;} return true;}
static boolean pal(String s){int n = s.length(),i1=0,i2=n-1;while(i1<i2){if(s.charAt(i1)!=s.charAt(i2))return false;i1++; i2--;}return true;}
 
// Input OutPut Functions
static PrintWriter out = new PrintWriter(System.out);
static FastReader sc=new FastReader();
static int ni(){ int x = sc.nextInt();return(x); }
static long nl(){ long x = sc.nextLong(); return(x); }
static String n(){ String str = sc.next(); return(str); }
static String ns(){ String str = sc.nextLine(); return(str);}
static double nd(){double d = sc.nextDouble(); return(d); }
static int[] ai(int n){ int a[] = new int[n]; for(int i=0;i<n;i++) a[i] = ni(); return a; }
static long[] al(int n) {long a[] = new long[n]; for(int i=0;i<n;i++) a[i] = nl(); return a; }
static double[] ad(int n) { double a[]= new double[n]; for(int i=0;i<n;i++) a[i]=nd(); return a;} 
static String[] as(int n){ String a[] = new String[n]; for(int i=0;i<n;i++) a[i] = n(); return a; }
static void p(Object o){ out.print(o); }
static void pn(Object o){ out.println(o); } 
 
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;} 
  } 
}