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;
const long long mod = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename A, typename B>
istream& operator>>(istream& input, pair<A, B>& x) {
  input >> x.first >> x.second;
  return input;
}
template <typename A>
istream& operator>>(istream& input, vector<A>& x) {
  for (auto& i : x) input >> i;
  return input;
}
template <typename A>
ostream& operator<<(ostream& output, vector<A>& x) {
  for (auto& i : x) output << i << ' ';
  return output;
}
void add(long long&, long long);
long long sub(long long, long long);
long long mul(long long, long long);
long long powz(long long, long long);
long long binpow(long long, long long);
long long gcd(long long, long long);
bool sortinrev(const pair<char, long long> a, pair<char, long long> b) {
  return (a.second > b.second);
}
const long long N = 1e9;
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  cin >> v;
  for (long long i = 0; i < n; i++) {
    if (v[i] > n) {
      cout << -1;
      return;
    }
  }
  set<long long> st;
  for (long long i = 1; i <= n; i++) {
    st.insert(i);
  }
  vector<long long> ans;
  ans.push_back(v[0]);
  st.erase(v[0]);
  for (long long i = 1; i < n; i++) {
    long long tmp = max(v[i], v[i - 1]);
    if (st.find(tmp) != st.end()) {
      st.erase(tmp);
      ans.push_back(tmp);
    } else {
      auto j = st.lower_bound(v[i]);
      if (j != st.begin()) --j;
      if (*j > tmp) {
        ans.push_back(tmp);
        continue;
      }
      ans.push_back(*j);
      st.erase(*j);
    }
  }
  vector<long long> w = ans;
  sort(w.begin(), w.end());
  for (long long i = 0; i < n - 1; i++) {
    if (w[i] == w[i + 1]) {
      cout << -1;
      return;
    }
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long tc = 1;
  cin >> tc;
  for (long long _ = 0; _ < tc; _++) {
    solve();
    if (_ != tc - 1) cout << "\n";
  }
}
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * 1ll * a;
    b >>= 1;
  }
  return res;
}
long long powz(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    b /= 2;
    a = mul(a, a);
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) {
    a += mod;
  }
  return a;
}
long long mul(long long a, long long b) { return ((a)*1ll * (b)) % mod; }
void add(long long& a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}