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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int bnm(long long int n, long long int k) {
  long long int res = 1;
  if (n < k) return 0;
  if (k > n - k) k = n - k;
  for (long long int i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long int power(long long int base, long long int exp) {
  long long int res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % 1000000007;
    base = (base * base) % 1000000007;
    exp /= 2;
  }
  return res % 1000000007;
}
long long int str(string s, char e) {
  for (long long int i = s.size() - 1; i >= 0; i--) {
    if (s[i] == e) return i;
  }
}
void ghost() {
  long long int n, i, k = 0, f = 0, g = 0, e = 0, u = 0, d = 0, q = 0, l = 0,
                      y = 0, s = 0, x = 0;
  cin >> q;
  while (q--) {
    queue<long long int> z;
    cin >> n;
    l = 0;
    long long int a[n + 1], b[n + 1], c[n + 1];
    for (long long int i = 1; i < n + 1; i++) a[i] = 0;
    for (long long int i = 1; i < n + 1; i++) c[i] = 1;
    b[0] = 0;
    a[0] = 0;
    for (long long int i = 1; i < n + 1; i++) {
      cin >> b[i];
    }
    for (long long int i = 1; i < n + 1; i++) {
      if (b[i] != b[i - 1]) {
        a[i] = b[i];
        c[b[i]]--;
      }
    }
    for (long long int i = 1; i < n + 1; i++) {
      if (c[i] == 1) z.push(i);
    }
    for (long long int i = 1; i < n + 1; i++) {
      if (a[i] == 0) {
        a[i] = z.front();
        z.pop();
      }
    }
    d = a[0];
    for (long long int i = 1; i < n + 1; i++) {
      if (max(d, a[i]) != b[i]) {
        cout << "-1" << endl;
        l = 1;
        break;
      }
      f = max(a[i], a[i - 1]);
      d = max(f, d);
    }
    if (l != 1) {
      for (long long int i = 1; i < n + 1; i++) cout << a[i] << " ";
      cout << endl;
    }
  }
}
signed main() {
  int test = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  while (test--) ghost();
  return 0;
}