Problem: Permutation p is a sequence of integers p=[p_1, p_2, ..., p_n], consisting of n distinct (unique) positive integers between 1 and n, inclusive. For example, the following sequences are permutations: [3, 4, 1, 2], [1], [1, 2]. The following sequences are not permutations: [0], [1, 2, 1], [2, 3], [0, 1, 2].

The important key is in the locked box that you need to open. To open the box you need to enter secret code. Secret code is a permutation p of length n. 

You don't know this permutation, you only know the array q of prefix maximums of this permutation. Formally:

  * q_1=p_1, 
  * q_2=max(p_1, p_2), 
  * q_3=max(p_1, p_2,p_3), 
  * ... 
  * q_n=max(p_1, p_2,...,p_n). 



You want to construct any possible suitable permutation (i.e. any such permutation, that calculated q for this permutation is equal to the given array).

Input

The first line contains integer number t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.

The first line of a test case contains one integer n (1 ≤ n ≤ 10^{5}) — the number of elements in the secret code permutation p.

The second line of a test case contains n integers q_1, q_2, ..., q_n (1 ≤ q_i ≤ n) — elements of the array q for secret permutation. It is guaranteed that q_i ≤ q_{i+1} for all i (1 ≤ i < n).

The sum of all values n over all the test cases in the input doesn't exceed 10^5.

Output

For each test case, print:

  * If it's impossible to find such a permutation p, print "-1" (without quotes). 
  * Otherwise, print n distinct integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n). If there are multiple possible answers, you can print any of them. 

Example

Input


4
5
1 3 4 5 5
4
1 1 3 4
2
2 2
1
1


Output


1 3 4 5 2 
-1
2 1 
1 

Note

In the first test case of the example answer [1,3,4,5,2] is the only possible answer:

  * q_{1} = p_{1} = 1; 
  * q_{2} = max(p_{1}, p_{2}) = 3; 
  * q_{3} = max(p_{1}, p_{2}, p_{3}) = 4; 
  * q_{4} = max(p_{1}, p_{2}, p_{3}, p_{4}) = 5; 
  * q_{5} = max(p_{1}, p_{2}, p_{3}, p_{4}, p_{5}) = 5. 



It can be proved that there are no answers for the second test case of the example.

Solution (in JAVA):

Solution:
//I AM THE CREED
/* //I AM THE CREED
/* package codechef; // don't place package name! */
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.util.*;
import java.awt.Point;
public class Main{
    static long mod=1000000007;
    public static void main(String[] args) throws IOException 
    { 
        Scanner input=new Scanner(System.in);
        input.nextInt();
        while(input.hasNext()){
            int n=input.nextInt();
            int[] prefix=new int[n];
            int ptr=1;
            for(int i=0;i<n;i++){
                prefix[i]=input.nextInt();
            }
            StringBuilder sol=new StringBuilder();
            HashSet<Integer> added=new HashSet<>();
            boolean possible=true;
            for(int i=0;i<n;i++){
                if(i==0){
                    sol.append(prefix[i]+" ");
                    added.add(prefix[i]);
                    continue;
                }
                if(prefix[i]>prefix[i-1]){
                    sol.append(prefix[i]+" ");
                    added.add(prefix[i]);
                    continue;
                }
                if(prefix[i]==prefix[i-1] && prefix[i]==1){
                    possible=false;
                    continue;
                }
                while(ptr<=n && added.contains(ptr)){
                    ptr++;
                }
                if(ptr>n || ptr>prefix[i]){
                    possible=false;
                    continue;
                }
                sol.append(ptr+" ");
                added.add(ptr);
            }
            if(possible==false || added.size()<n){
                System.out.println(-1);
            }
            else{
                System.out.println(sol.toString());
            }
        }
        
    }
    static long pow(long a, long b){
        if(b==1){
            return a%mod;
        }
        return (pow(a, b-1)*a)%mod;
    }
    static long dfs(ArrayList<ArrayList<Integer>> g, HashSet<Integer> v, int n, HashMap<String, Integer> col){
        if(v.contains(n)){
            return (long)(0);
        }
        long tot=1;
        v.add(n);
        for(int children:g.get(n)){
            if(col.get(Integer.toString(n)+","+Integer.toString(children))==1){
                continue;
            }
            tot+=dfs(g, v, children, col);
        }
        return tot;
    }
    //Credits to SecondThread(https://codeforces.com/profile/SecondThread) for this tempelate
    static void ruffle_sort(long[] a) {
		// shandom_ruffle
		Random r=new Random();
		int n=a.length;
		for (int i=0; i<n; i++) {
			int oi=r.nextInt(i);
			long temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
 
		// sort
		Arrays.sort(a);
	}
    //Credits to SecondThread(https://codeforces.com/profile/SecondThread) for this tempelate.
    static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
    
    
}