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;
const long long MOD = 1e9 + 7;
void printv(vector<long long> arr, long long n) {
  for (long long i = 0; i < n; ++i) cout << arr[i] << " ";
  cout << endl;
}
void print(long long arr[], long long n) {
  for (long long i = 0; i < n; ++i) cout << arr[i] << " ";
  cout << endl;
}
bool isPrime(long long n) {
  bool flag = 1;
  for (long long i = 2; i <= sqrt(n); ++i) {
    if (n % i == 0) {
      flag = 0;
      break;
    }
  }
  return (flag ? 1 : 0);
}
long long mod_power(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % MOD;
    x = (x * x) % MOD;
    y >>= 1;
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    vector<long long> a(n), vis(n + 1, 0), res(n, -1), rem;
    for (long long i = 0; i < n; ++i) {
      cin >> a[i];
    }
    bool ok = true;
    for (long long i = 0; i < n; ++i) {
      if (a[i] < i + 1) {
        ok = false;
        break;
      }
    }
    if (!ok) {
      cout << "-1\n";
      continue;
    }
    for (long long i = 0; i < n; ++i) {
      if (!vis[a[i]]) {
        res[i] = a[i];
        vis[a[i]] = 1;
      }
    }
    for (long long i = 1; i <= n; ++i) {
      if (!vis[i]) rem.push_back(i);
    }
    vis.clear();
    long long j = 0;
    for (long long i = 0; i < n; ++i) {
      if (res[i] == -1) {
        res[i] = rem[j++];
      }
    }
    printv(res, n);
  }
  return 0;
}