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>
#pragma GCC optimize("O3")
using namespace std;
long double eps = 1.0 / 1e18;
long long sqr(long long a) { return a * a; };
long long gcd(long long a, long long b) { return !a ? b : gcd(b % a, a); }
long long binpow(long long a, long long b, long long mod) {
  return b ? (b % 2 ? (a * (sqr(binpow(a, b / 2, mod)) % mod)) % mod
                    : sqr(binpow(a, b / 2, mod)) % mod)
           : 1;
}
long long a[100001], ans[100001];
int main() {
  cout.tie(0), cin.tie(0), ios_base::sync_with_stdio(0);
  long long q, n, i;
  cin >> q;
  while (q--) {
    cin >> n;
    bool bl = 0;
    set<long long> st;
    for (i = 1; i <= n; i++) cin >> a[i], ans[i] = 0, st.insert(i);
    for (i = 1; i <= n; i++) {
      if (i == 1)
        ans[1] = a[1], st.erase(a[1]);
      else {
        if (a[i] > a[i - 1]) {
          if (st.find(a[i]) == st.end())
            bl = 1;
          else
            ans[i] = a[i], st.erase(a[i]);
        }
      }
    }
    for (i = 1; !bl && i <= n && !st.empty(); i++) {
      if (ans[i]) continue;
      auto f = *(st.begin());
      if (f < a[i])
        ans[i] = f, st.erase(f);
      else {
        bl = 1;
        break;
      }
    }
    for (i = 1; !bl && i <= n; i++)
      if (!ans[i]) bl = 1;
    if (!bl) {
      for (i = 1; i <= n; i++) cout << ans[i] << ' ';
    } else
      cout << -1;
    cout << '\n';
  }
  return 0;
}