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;
bool isPrime(int x) {
  if (x == 1) {
    return false;
  }
  if (x == 2 || x == 3) {
    return true;
  }
  if (x % 2 == 0 || x % 3 == 0) {
    return false;
  }
  for (int i = 5; i * i <= x; i += 6) {
    if (x % i == 0 || x % (i + 2) == 0) {
      return false;
    }
  }
  return true;
}
long long gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(int a, int b) { return (a / gcd(a, b)) * b; }
void factorize(vector<pair<int, int>> &v, int num) {
  for (int i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      int cnt = 0;
      while (num % i == 0) {
        cnt++;
        num /= i;
      }
      v.push_back({i, cnt});
    }
  }
  if (num > 1) {
    v.push_back({num, 1});
  }
}
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
long long int power(long long int a, long long int n, long long int p) {
  long long int res = 1;
  while (n) {
    if (n % 2) {
      res = ((res % p) * (a % p)) % p;
    }
    a = ((a % p) * (a % p)) % p;
    n /= 2;
  }
  return res;
}
void sieve() {
  int arr[1000001];
  int maxN = 1000000;
  for (int i = 1; i <= maxN; i++) {
    arr[i] = -1;
  }
  for (int i = 2; i <= maxN; i++) {
    if (arr[i] == -1) {
      for (int j = i; j <= maxN; j += i) {
        if (arr[j] == -1) {
          arr[j] = i;
        }
      }
    }
  }
  int n = 100;
  cout << arr[n];
  n /= arr[n];
  while (n > 1) {
    cout << " x " << arr[n];
    n /= arr[n];
  }
}
long long int ncr(long long int n, long long int r) {
  long long int res = 1;
  if (r > n - r) {
    r = n - r;
  }
  for (long long int i = 0; i < r; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long int fact(long long int n) {
  long long int fac = 1;
  for (int i = 0; i < n; i++) {
    fac = fac * (i + 1);
  }
  return fac;
}
void show(vector<int> ans) {
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
}
int decode() {
  int n;
  cin >> n;
  int arr[n];
  vector<int> v;
  for (int i = 0; i < n; i++) {
    v.push_back(i + 1);
  }
  unordered_set<int> s;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  int smallest = 1;
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    if (i == 0) {
      s.insert(arr[i]);
      ans.push_back(arr[i]);
      v.erase(lower_bound(v.begin(), v.end(), arr[i]));
      continue;
    }
    if (s.find(arr[i]) == s.end()) {
      v.erase(lower_bound(v.begin(), v.end(), arr[i]));
      s.insert(arr[i]);
      ans.push_back(arr[i]);
    } else {
      auto it = upper_bound(v.begin(), v.end(), arr[i]);
      if (it != v.begin()) {
        ans.push_back(*v.begin());
        v.erase(v.begin());
      } else {
        cout << -1 << "\n";
        return 0;
      }
    }
  }
  show(ans);
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    decode();
  }
  return 0;
}