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 maxn = 3e5 + 20;
const int inf = 0x3f3f3f3f;
const int modd = 1e9 + 7;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f * x;
}
template <class T>
inline void sc(T &x) {
  char c;
  x = 0;
  while ((c = getchar()) < '0')
    ;
  while (c >= '0' && c <= '9') x = x * 10 + (c - 48), c = getchar();
}
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
inline long long exgcd(long long a, long long b, long long &x, long long &y) {
  long long d;
  (b == 0 ? (x = 1, y = 0, d = a)
          : (d = exgcd(b, a % b, y, x), y -= a / b * x));
  return d;
}
inline long long qpow(long long a, long long n) {
  long long sum = 1;
  while (n) {
    if (n & 1) sum = sum * a % modd;
    a = a * a % modd;
    n >>= 1;
  }
  return sum;
}
inline long long qmul(long long a, long long n) {
  long long sum = 0;
  while (n) {
    if (n & 1) sum = (sum + a) % modd;
    a = (a + a) % modd;
    n >>= 1;
  }
  return sum;
}
inline long long inv(long long a) { return qpow(a, modd - 2); }
struct Coke {
  int v, id;
} a[maxn];
bool cmp1(Coke c, Coke d) { return c.v < d.v; }
bool cmp2(Coke c, Coke d) { return c.id < d.id; }
int vis[205];
int ans[maxn];
vector<int> vec;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= 2 * n; i++) vis[i] = 0;
    vec.clear();
    for (int i = 1; i <= n; i++)
      scanf("%d", &a[i].v), a[i].id = i, ans[i * 2 - 1] = a[i].v;
    sort(a + 1, a + 1 + n, cmp1);
    for (int i = 1; i <= n; i++) vis[a[i].v] = a[i].id;
    int now = 0, flag = 1;
    for (int i = 1; i <= 2 * n; i++)
      if (!vis[i]) vec.push_back(i);
    sort(a + 1, a + 1 + n, cmp2);
    for (int i = 1; i <= n; i++) {
      auto it = lower_bound(vec.begin(), vec.end(), a[i].v);
      if (it == vec.end()) {
        flag = 0;
        break;
      }
      ans[a[i].id * 2] = *it;
      vec.erase(it);
    }
    if (!flag) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= 2 * n; i++) printf("%d%c", ans[i], " \n"[i == 2 * n]);
  }
}