Problem: You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.

Input

Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).

The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).

The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).

It is guaranteed that the sum of n by all test cases doesn't exceed 100.

Output

For each test case, if there is no appropriate permutation, print one number -1.

Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.

Example

Input


5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8


Output


1 2 
-1
4 5 1 2 3 6 
-1
1 3 5 6 7 9 2 4 8 10 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int powmod(int a, int b, int m) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * 1ll * a) % m;
    }
    a = (a * 1ll * a) % m;
    b >>= 1;
  }
  return res;
}
int discrete_log(int a, int b, int m) {
  b %= (m - 1);
  int n = (int)sqrt(m + .0) + 1;
  map<int, int> vals;
  for (int p = n; p >= 1; --p) vals[powmod(a, p * n, m)] = p;
  for (int q = 0; q <= n; ++q) {
    int cur = (powmod(a, q, m) * 1ll * b) % m;
    if (vals.count(cur)) {
      int ans = vals[cur] * n - q;
      return ans;
    }
  }
  return -1;
}
int generate_primitive_root(int p) {
  vector<int> fact;
  int phi = p - 1, n = phi;
  for (int i = 2; i * i <= n; ++i)
    if (n % i == 0) {
      fact.push_back(i);
      while (n % i == 0) n /= i;
    }
  if (n > 1) fact.push_back(n);
  for (int res = 2; res <= p; ++res) {
    bool ok = true;
    for (size_t i = 0; i < fact.size() && ok; ++i)
      ok &= powmod(res, phi / fact[i], p) != 1;
    if (ok) return res;
  }
  return -1;
}
vector<vector<int>> mat_mul(vector<vector<int>> a, vector<vector<int>> b) {
  int N = a.size();
  vector<vector<int>> ans(N, vector<int>(N, 0));
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      for (int k = 0; k < N; k++) {
        ans[i][j] += a[i][k] + a[k][j];
        ans[i][j] %= (int)(1e9 + 7);
      }
    }
  }
  return ans;
}
int main() {
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    vector<int> a(2 * n + 2);
    set<int> st;
    for (int i = 1; i <= 2 * n; i++) st.insert(i);
    for (int i = 1; i <= n; i++) {
      cin >> a[2 * i - 1];
      st.erase(a[2 * i - 1]);
    }
    bool ok = true;
    for (int i = 1; i <= n; i++) {
      auto it = st.upper_bound(a[2 * i - 1]);
      if (it == st.end()) {
        ok = false;
        break;
      } else {
        int val = *it;
        st.erase(val);
        a[2 * i] = val;
      }
    }
    if (!ok)
      cout << -1 << endl;
    else {
      for (int i = 1; i <= 2 * n; i++) cout << a[i] << " ";
      cout << endl;
    }
  }
  return 0;
}