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:

import java.util.*;
import java.io.*;
public class Codeforces6 {

	public static void main(String[] args) throws IOException{
		// TODO Auto-generated method stub
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int t=Integer.parseInt(br.readLine());
		while(t--!=0) {
			int n=Integer.parseInt(br.readLine());
			String s[]=br.readLine().split(" ");
			int a[]=new int[n];
			for(int i=0;i<n;i++) a[i]=Integer.parseInt(s[i]);
			int ch=1;
			ArrayList<Integer> ans=new ArrayList<>();
			ArrayList<Integer> v=new ArrayList<>();
			Set<Integer> hs=new HashSet<>();
			if(n==1) {
				if(a[0]==1) System.out.println('1');
				else System.out.println("-1");
				continue;
			}
			ans.add(a[0]);
			hs.add(a[0]);
			if(!hs.contains(ch)) v.add(ch);
			boolean ok=true;
			for(int i=1;i<n;i++) {
				if(a[i]==a[i-1]) {
					while(v.size()!=0 && hs.contains(v.get(v.size()-1))) {
						v.remove(v.size()-1);
					}
					if(v.size()==0) {
						ok=false;
						break;
					}
					
					ans.add(v.get(v.size()-1));
					hs.add(v.get(v.size()-1));
				}
				else {
					ans.add(a[i]);
					hs.add(a[i]);
				}
				ch++;
				if(!hs.contains(ch)) v.add(ch);
			}
			if(!ok) {
				System.out.println("-1");
				continue;
			}
			for(Integer c:ans) System.out.print(c+" ");
			System.out.println();
		}
	}
}