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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long pows(long long b, long long e) {
  if (e == 0)
    return 1;
  else if (e % 2 == 0) {
    long long a = pow(b, e / 2);
    return a * a;
  } else {
    long long a = pow(b, e / 2);
    return b * a * a;
  }
}
long long powm(long long x, long long y, long long m) {
  x = x % m;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= m;
    y = y >> 1;
    x = x * x;
    x %= m;
  }
  return res;
}
long long modInverse(long long a, long long m = 998244353) {
  if (m == 1) return 0;
  long long m0 = m, y = 0, x = 1;
  while (a > 1) {
    long long q = a / m, t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    long long n, m, i, j, k, x, y, z, e, f, p, q, g, l, r, w, h,
        count1 = 0, prod = 1, a, b, c, d, index, x1, x2, diff, ans = 0, sum = 0,
        sum1 = 0, sum2 = 0, flag = 0, flag1 = 0, flag2 = 0;
    string s, s1, s2;
    cin >> n;
    long long ar[n];
    for (i = 0; i < n; i++) {
      cin >> ar[i];
    }
    set<long long> st;
    for (i = 1; i <= n; i++) {
      st.insert(i);
    }
    vector<long long> v;
    long long maxi = ar[0];
    v.push_back(ar[0]);
    unordered_map<long long, long long> m1;
    m1[ar[0]] = 1;
    st.erase(ar[0]);
    for (i = 1; i < n; i++) {
      if (ar[i] > maxi) {
        maxi = ar[i];
        st.erase(maxi);
        v.push_back(ar[i]);
      } else {
        p = *st.begin();
        if (p >= ar[i]) {
          flag = 1;
          break;
        } else {
          st.erase(p);
          v.push_back(p);
        }
      }
    }
    if (flag) {
      cout << "-1\n";
    } else {
      for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
      ;
      cout << "\n";
      ;
    }
  }
}