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;
int power(int x, int y, int p) {
  int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int logN(int base, int num) {
  if (base == 1 && num != 1) return -1;
  if (num / base == 0) return 0;
  return 1 + logN(base, num / base);
}
bool isPrime(long long 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;
}
bool perfect(long long n) {
  return (long long)sqrt(n) * (long long)sqrt(n) == n ? true : false;
}
bool sortcol(const vector<int>& v1, const vector<int>& v2) {
  return v1[0] < v2[0];
}
int main() {
  int t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    int pos = 1, arr[n], ans[n], flag[n + 1];
    vector<int> v;
    for (int i = 0; i < (n); i++) cin >> arr[i], flag[i + 1] = 1;
    for (int i = n; i >= 1; i--) v.push_back(i);
    for (int i = 0; i < (n); i++) {
      if (i > 0 && arr[i] == arr[i - 1]) {
        while (!v.empty() && !flag[v.back()]) v.pop_back();
        if (v.empty() || v.back() > arr[i]) {
          pos = 0;
          break;
        } else {
          ans[i] = v.back();
          v.pop_back();
        }
      } else
        ans[i] = arr[i], flag[arr[i]] = 0;
    }
    if (pos)
      for (int i = 0; i < (n); i++) cout << ans[i] << ' ';
    else
      cout << -1;
    cout << "\n";
  }
  return 0;
}