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:
//package godiji;

import java.util.Scanner;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.PriorityQueue;

public class pac1 {
	static Scanner scn = new Scanner(System.in);
	static int mod = 1000000007;

	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;
		}
	}

	static FastReader s = new FastReader();

	public static void main(String[] args) {
		int t = scn.nextInt();
		while (t-- > 0) 
		{
			int n = scn.nextInt();
			int arr[]=new int[n];
			HashMap<Integer, Integer>map=new HashMap<Integer, Integer>();
			for(int i=0;i<n;i++)
			{
				arr[i]=scn.nextInt();
				if(map.containsKey(arr[i]))
					map.put(arr[i],map.get(arr[i])+1);
				else
					map.put(arr[i],1);
			}
			boolean value=true;
			for(int val:map.keySet())
			{
				if(map.get(val)>val)
				{
					value=false;
					break;
				}
			}
			boolean bl=true;
			for(int i=0;i<n-1;i++)
			{
				if(arr[i]>arr[i+1])
				{
					bl=false;
					break;
				}
			}
			if(bl&&value)
			{
				int p[]=new int[n],min=1;
				p[0]=arr[0];
				map.clear();
				if(arr[0]==min)
					min++;
				map.put(arr[0],1);
				for(int i=1;i<n;i++)
				{
					if(arr[i]==arr[i-1])
					{
						if(!map.containsKey(min))
						{	
							map.put(min, 1);
							p[i]=min;
							min++;
						}
						else
						{
							while(map.containsKey(min))
							{
								min++;
							}
							map.put(min, 1);
							p[i]=min;
							min++;
						}
					}
					else
					{
						p[i]=arr[i];
						map.put(arr[i],1);
					}
				}
				boolean tr=true;
				
				for(int i=0;i<n;i++)
				{
					if(p[i]>arr[i])
					{
						tr=false;
						break;
					}
				}
				if(tr)
				{
					for(int i=0;i<n;i++)
						System.out.print(p[i]+" ");
					System.out.println();
				}
				else
					System.out.println("-1");
			}
			else
				System.out.println("-1");
			
		}
	}

	public static boolean strong(String str) {
		int n = str.length();

		return true;
	}

	public static boolean[] sieveOfEratosthenes(int n) {

		boolean prime[] = new boolean[n + 1];
		for (int i = 0; i < n; i++)
			prime[i] = true;

		for (int p = 2; p * p <= n; p++) {
			// If prime[p] is not changed, then it is a prime
			if (prime[p] == true) {
				// Update all multiples of p
				for (int i = p * p; i <= n; i += p)
					prime[i] = false;
			}
		}

		for (int i = 2; i <= n; i++) {
			if (prime[i] == true) {
				// list.add(i);
			}

		}
		return prime;
	}

	public static long string(int n, String str, long memo[]) {
		if (str.length() == n) {
			boolean bl = true;
			for (int i = 0; i < n - 1; i++) {
				if (str.charAt(i) == '1' && str.charAt(i + 1) == '1') {
					bl = false;
					break;
				}
			}
			if (bl)
				return 1;
			else
				return 0;
		}

		long a = (string(n, str + "0", memo)) % mod;
		long b = (string(n, str + "1", memo)) % mod;
		return (a + b) % mod;
	}

	public static long gcd(long a, long n) {

		if (a == 0)
			return n;
		return gcd(n % a, a);
	}

	public static long lcm(long a, long b) {
		long s = (a * b) / gcd(a, b);
		return s;
	}
}

class pair {
	int a;
	int b;

	pair(int fir, int sec) {
		this.a = fir;
		this.b = sec;
	}
}

class sorting implements Comparator<pair> {

	@Override
	public int compare(pair o1, pair o2) {
		// TODO Auto-generated method stub
		int diff = o1.a - o2.a;
		if (diff > 0)
			return 1;
		else
			return -1;
	}

}