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;
void vec(vector<int>& v) {
  for (int i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << endl;
}
void vec_ll(vector<long long int>& v) {
  for (int i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << endl;
}
int parsestring(string temp) {
  int num = 0;
  int digs = temp.size() - 1;
  for (int i = 0; i <= digs; i++) {
    num += pow(10, digs - i) * (temp[i] - '0');
  }
  return num;
}
bool prime(long long int n) {
  if (n < 2) return false;
  for (long long int x = 2; x * x <= n; x++) {
    if (n % x == 0) return false;
  }
  return true;
}
string ter(long long int n) {
  string t = "";
  while (n > 0) {
    t += to_string(n % 3);
    n /= 3;
  }
  return t;
}
long long int power(int base, int exp) {
  long long int ans = 1;
  for (int i = 1; i <= exp; i++) {
    ans *= base;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t) {
    int n;
    cin >> n;
    vector<int> q(n);
    for (int i = 0; i < n; i++) {
      cin >> q[i];
    }
    vector<int> a(n);
    vector<bool> visit(n + 1, false);
    int pmax = -1;
    int ptr = 1;
    bool inc = true;
    for (int i = 0; i < n; i++) {
      if (pmax != q[i]) {
        if (q[i] >= pmax) {
          a[i] = q[i];
          pmax = q[i];
          visit[q[i]] = true;
        } else {
          inc = false;
          break;
        }
      } else {
        while (visit[ptr]) {
          ptr++;
        }
        if (ptr < pmax) {
          a[i] = ptr;
          visit[ptr] = true;
          ptr++;
        } else {
          inc = false;
          break;
        }
      }
    }
    if (!inc) {
      cout << -1 << endl;
    } else {
      vec(a);
    }
    t--;
  }
}