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;
const int N = (int)100;
const int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};
char dir[] = {'L', 'D', 'R', 'U'};
const long long int mod = (long long int)1e9 + 7;
const long long int inf = 1e18 + 100;
int BIT[100005], n;
long long int prime[17], spf[N];
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <typename T>
T min3(T a, T b, T c) {
  return min(a, min(b, c));
}
template <typename T>
T max3(T a, T b, T c) {
  return max(a, max(b, c));
}
bool isVowel(char ch) {
  ch = toupper(ch);
  if (ch == 'A' || ch == 'U' || ch == 'I' || ch == 'O' || ch == 'E')
    return true;
  return false;
}
long long int mInv(long long int a, long long int m) {
  long long int m0 = m;
  long long int y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long int q = a / m;
    long long int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long Extendedgcd(long long a, long long b, long long *x, long long *y) {
  if (!a) {
    *x = 0;
    *y = 1;
    return b;
  }
  long long nx, ny;
  long long d = Extendedgcd(b % a, a, &nx, &ny);
  long long k = b / a;
  *y = nx;
  *x = -k * nx + ny;
  return d;
}
void update(int x, int delta) {
  if (!x) {
    BIT[x] += delta;
    return;
  }
  for (; x <= N; x += x & -x) BIT[x] += delta;
}
long long int query(int x) {
  long long int sum = BIT[0];
  for (; x > 0; x -= x & -x) sum += BIT[x];
  return sum;
}
void Sieve() {
  for (int i = 1; i <= N; i++) prime[i] = 1;
  prime[0] = 0;
  prime[1] = 0;
  for (long long int p = 2; p * p <= N; p++) {
    if (prime[p] == 1) {
      for (long long int i = p * p; i <= N; i += p) prime[i] = 0;
    }
  }
  for (int p = 3; p <= N; p++) prime[p] += prime[p - 1];
}
void sieve() {
  spf[1] = -1;
  for (int i = 2; i < N; i++) spf[i] = i;
  for (int i = 4; i < N; i += 2) spf[i] = 2;
  for (long long int i = 3; i * i < N; i++) {
    if (spf[i] == i) {
      for (long long int j = i * i; j < N; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
struct cmp {
  bool operator()(const pair<int, int> &x, const pair<int, int> &y) const {
    if (x.first != y.first) return x.first < y.first;
    return x.second > y.second;
  }
};
void gInput(int m, vector<int> adj[]) {
  int u, v;
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
}
vector<long long int> fac;
void factorial() {
  int N = (int)1e6;
  fac.resize(N);
  fac[0] = 1;
  for (long long int i = 1; i <= N; i++) {
    fac[i] = ((fac[i - 1] % mod) * i) % mod;
  }
}
long long int nCr(long long int n, long long int r) {
  long long int x = fac[n];
  long long int y = fac[r];
  long long int z = fac[n - r];
  long long int pr = ((y % mod) * (z % mod)) % mod;
  pr = mInv(pr, mod);
  long long int ncr = ((x % mod) * (pr % mod)) % mod;
  return ncr;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int a[n + 1];
    int b[2 * n + 1];
    int vis[2 * n + 1];
    memset(vis, 0, sizeof(vis));
    memset(b, 0, sizeof(b));
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      vis[a[i]] = 1;
      b[2 * i - 1] = a[i];
    }
    int f = 0;
    for (int i = 1; i <= 2 * n; i++) {
      if (b[i] == 0) {
        int g = 0;
        for (int j = 1; j <= 2 * n; j++) {
          if (!vis[j] and j > b[i - 1]) {
            b[i] = j;
            vis[j] = 1;
            g = 1;
            break;
          }
        }
        if (!g) f = 1;
      }
    }
    if (f)
      cout << -1;
    else
      for (int i = 1; i <= 2 * n; i++) cout << b[i] << " ";
    cout << "\n";
  }
  return 0;
}