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>
const int N = 1e6;
const double eps = 0.00000001;
const long long mod = 1e9 + 7;
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int a[1000], b[1000];
bool vis[1000];
int main() {
  int t = read();
  while (t--) {
    memset(vis, 0, sizeof vis);
    memset(b, 0, sizeof b);
    int n = read();
    bool r = 1;
    for (int i = 1; i <= n; ++i) {
      a[i] = read(), vis[a[i]] = 1;
      if (a[i] == 2 * n) r = 0;
    }
    if (r == 0) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; ++i) {
      b[2 * i - 1] = a[i];
      r = 1;
      for (int j = a[i]; j <= 2 * n; ++j)
        if (!vis[j]) {
          b[2 * i] = j;
          vis[j] = 1;
          r = 0;
          break;
        }
      if (r == 1) break;
    }
    r = 1;
    for (int i = 1; i <= 2 * n; ++i)
      if (b[i] == 0) {
        r = 0;
        break;
      }
    if (r == 0)
      puts("-1");
    else {
      for (int i = 1; i <= 2 * n; ++i) cout << b[i] << ' ';
      cout << endl;
    }
  }
}