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.lang.*;
import java.io.*;
import java.math.*;


public class file{


// *********************************************BEGINS*********************************************
public static void main(String[] args) throws IOException{


// Working program using BufferedReader 
        //BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); 
       // StringTokenizer st = new StringTokenizer(br.readLine()); 
       // int n = Integer.parseInt(st.nextToken()); 
       // int k = Integer.parseInt(st.nextToken()); 

Reader scn=new Reader();
PrintWriter pt=new PrintWriter(System.out);

int test=scn.nextInt();
while(test-->0) {
	int n=scn.nextInt();
HashMap<Integer,Integer> co=new HashMap<>();
HashMap<Integer,Integer> left=new HashMap<>();
	
	int[] arr=new int[n];
	for(int i=0;i<n;i++) {
		arr[i]=scn.nextInt();
	}

HashSet<Integer> hs=new HashSet<>();
int[] ans=new int[n];
ans[0]=arr[0];
hs.add(ans[0]);

boolean flag=false;
for(int i=1;i<n;i++) {
	if(arr[i]!=arr[i-1]) {
		ans[i]=arr[i];
		hs.add(arr[i]);
	}
}
int ind=0;
for(int i=1;i<=n;i++) {
	if(hs.contains(i)) {
		continue;
	}else {
		while(ans[ind]!=0) {
			ind++;
		}
		ans[ind++]=i;
	}
}
for(int i=0;i<n;i++) {
	if(ans[i] > arr[i]) {
		flag=true;
		break;
		
	}
}
	if(flag) {
		pt.print("-1");
	}else {
		for(int i:ans)
			pt.print(i+" ");
	}
	pt.println("");
}


pt.close();

}
// *********************************************ENDED*********************************************
static int abs(int x) {
		return Math.abs(x);
	}
 
static long abs(long x) {
		return Math.abs(x);
	}

static int max(int x, int y) {
		return Math.max(x, y);
	}
 
static int min(int x, int y) {
		return Math.min(x, y);
	}
 
static long max(long x, long y) {
		return Math.max(x, y);
	}
 
static long min(long x, long y) {
		return Math.min(x, y);
	}

static int min(int a[]) {
		int x = 1_000_000_00_9;
		for (int i = 0; i < a.length; i++)
			x = min(x, a[i]);
		return x;
	}
 
static int max(int a[]) {
		int x = -1_000_000_00_9;
		for (int i = 0; i < a.length; i++)
			x = max(x, a[i]);
		return x;
	}
 
static long min(long a[]) {
		long x = (long) 3e18;
		for (int i = 0; i < a.length; i++)
			x = min(x, a[i]);
		return x;
	}
 
static long max(long a[]) {
		long x = -(long) 3e18;
		for (int i = 0; i < a.length; i++)
			x = max(x, a[i]);
		return x;
	}

static int power(int x, int y) {
		int res = 1;
		while (y > 0) {
			if (y % 2 == 1)
				res = (res * x);
			y >>= 1;
			x = (x * x);
		}
		return res;
	}

static long power(long x, long y) {
		long res = 1;
		while (y > 0) {
			if (y % 2 == 1)
				res = (res * x);
			y >>= 1;
			x = (x * x);
		}
		return res;
	}
	
static boolean isPrime(long n) 
{ 
    // Converting long to BigInteger 
    BigInteger b = new BigInteger(String.valueOf(n)); 

    return b.isProbablePrime(1); 
} 

static void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];
 
        /*Copy data to temp arrays*/
        for (int i = 0; i < n1; ++i)
            L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j)
            R[j] = arr[m + 1 + j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
    static void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l + r) / 2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m + 1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

static class Reader 
    { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead; 
  

public Reader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        }


public Reader(String file_name) throws IOException 
        { 
            din = new DataInputStream(new FileInputStream(file_name)); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        }  

public String readLine() throws IOException 
        { 
            byte[] buf = new byte[64]; // line length 
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        } 


 public int nextInt() throws IOException 
        { 
            int ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do
            { 
                ret = ret * 10 + c - '0'; 
            }  while ((c = read()) >= '0' && c <= '9'); 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 


 public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        } 


public double nextDouble() throws IOException 
        { 
            double ret = 0, div = 1; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
  
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
  
            if (c == '.') 
            { 
                while ((c = read()) >= '0' && c <= '9') 
                { 
                    ret += (c - '0') / (div *= 10); 
                } 
            } 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 

private void fillBuffer() throws IOException 
        { 
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
            if (bytesRead == -1) 
                buffer[0] = -1; 
        } 

private byte read() throws IOException 
        { 
            if (bufferPointer == bytesRead) 
                fillBuffer(); 
            return buffer[bufferPointer++]; 
        } 


public void close() throws IOException 
        { 
            if (din == null) 
                return; 
            din.close(); 
        } 
    }
  
}