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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class rand {

    static final long mod = (long) 1e9 + 7l;

    private static void solve(int t) {
        int n = fs.nextInt();
        boolean arr[] = new boolean[n + 1];
        int q[] = fs.readArray(n);
        int ans[] = new int[n];
        int k = 1;
        int lastnum = -1;
        for (int i = 1; i <= n; i++) {
            int num = q[i - 1];
            if (num < i) {
                out.println(-1);
                return;
            }
            if (i == 1 || num > lastnum) {
                ans[i - 1] = num;
                arr[num] = true;
                lastnum = num;
            } else {
                while (arr[k] == true) {
                    k++;
                }
                ans[i - 1] = k;
                arr[k] = true;
            }
        }
        for (int i = 0; i < n; i++) {
            out.print(ans[i] + " ");
        }
        out.println();

    }

    public static void main(String[] args) {
        fs = new FastScanner();
        out = new PrintWriter(System.out);
        long s = System.currentTimeMillis();
        int t = fs.nextInt();
        for (int i = 1; i <= t; i++) solve(t);
        out.close();
//        System.err.println( System.currentTimeMillis() - s + "ms" );
    }

    static boolean DEBUG = true;
    static PrintWriter out;
    static FastScanner fs;

    static void trace(Object... o) {
        if (!DEBUG) return;
        System.err.println(Arrays.deepToString(o));
    }

    static void pl(Object o) {
        out.println(o);
    }

    static void p(Object o) {
        out.print(o);
    }

    static long gcd(long a, long b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static int gcd(int a, int b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static void sieveOfEratosthenes(int n, int factors[]) {
        factors[1] = 1;
        for (int p = 2; p * p <= n; p++) {
            if (factors[p] == 0) {
                factors[p] = p;
                for (int i = p * p; i <= n; i += p)
                    factors[i] = p;
            }
        }
    }

    static long mul(long a, long b) {
        return a * b % mod;
    }

    static long fact(int x) {
        long ans = 1;
        for (int i = 2; i <= x; i++) ans = mul(ans, i);
        return ans;
    }

    static long fastPow(long base, long exp) {
        if (exp == 0) return 1;
        long half = fastPow(base, exp / 2);
        if (exp % 2 == 0) return mul(half, half);
        return mul(half, mul(half, base));
    }

    static long modInv(long x) {
        return fastPow(x, mod - 2);
    }

    static long nCk(int n, int k) {
        return mul(fact(n), mul(modInv(fact(k)), modInv(fact(n - k))));
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        public String next() {
            while (!st.hasMoreElements())
                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());
        }
    }
}