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 int mod = 1e9 + 7;
template <typename X>
inline X abs(const X& a) {
  return a < 0 ? -a : a;
}
template <typename X>
inline X sqr(const X& a) {
  return a * a;
}
bool rev(pair<long long int, long long int>& a,
         pair<long long int, long long int>& b) {
  return a.first < b.first || (a.first == b.first && a.second > b.second);
}
int dx[4] = {-1, +1, 0, 0};
int dy[4] = {0, 0, +1, -1};
int DX[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int DY[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
const int N = 2e5 + 5;
bool test() {
  int n;
  cin >> n;
  set<int> have;
  for (int i = 1; i <= n; ++i) {
    have.insert(i);
  }
  vector<int> q(n), p(n, 0);
  for (int i = 0; i < n; ++i) {
    cin >> q[i];
  }
  for (int i = 0; i < n; ++i) {
    if (i == 0 || q[i] > q[i - 1]) {
      if (!have.count(q[i])) {
        return false;
      }
      have.erase(q[i]);
      p[i] = q[i];
    }
  }
  for (int i = 0; i < n; ++i) {
    if (!p[i]) {
      p[i] = *have.begin();
      have.erase(have.begin());
      if (p[i] > q[i]) {
        return false;
      }
    }
  }
  for (int e : p) {
    cout << e << " ";
  }
  cout << "\n";
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) {
    if (!test()) {
      cout << "-1\n";
    }
  }
  return 0;
}