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 = 1000000007;
void print() { cout << endl; }
void printarray() { return; }
void scan() { return; }
template <typename T, typename... Args>
void print(T a, Args... arg) {
  cout << a << " ";
  print(arg...);
}
template <typename T, typename... Args>
void scan(T &a, Args &...arg) {
  cin >> a;
  scan(arg...);
}
template <typename T, typename... Args>
void scan(vector<T> &v, Args &...arg) {
  for (auto &i : v) cin >> i;
  scan(arg...);
}
template <typename T, typename... Args>
void printarray(vector<T> &v, Args &...arg) {
  for (auto i : v) cout << i << " ";
  cout << endl;
  printarray(arg...);
}
const long long int mxN = 1e6 + 5;
vector<long long int> prime(mxN);
void sieve() {
  for (long long int i = 0; i < mxN; i++) {
    prime[i] = true;
  }
  prime[0] = false;
  prime[1] = false;
  for (long long int p = 2; p * p < mxN; p++) {
    if (prime[p]) {
      for (long long int i = p * p; i < mxN; i += p) {
        prime[i] = false;
      }
    }
  }
}
long long int isprime(long long int x) {
  if (x <= 1) return 0;
  for (long long int i = 2; i <= x / 2; i++) {
    if (x % i == 0) {
      return 0;
    }
  }
  return 1;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
void stryker27() {
  long long int n;
  cin >> n;
  vector<long long int> a(n), dp;
  scan(a);
  set<long long int> s;
  for (long long int i = 1; i < n + 1; i++) s.insert(i);
  for (long long int i = 0; i < n; i++) {
    if (s.find(a[i]) != s.end()) {
      dp.push_back(a[i]);
      s.erase(a[i]);
    } else {
      long long int x = *s.begin();
      if (x > a[i]) {
        print(-1);
        return;
      }
      dp.push_back(x);
      s.erase(x);
    }
  }
  printarray(dp);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) stryker27();
  return 0;
}