C. Restoring Permutation

链接：https://codeforces.ml/contest/1315/problem/C

You are given a sequence b1,b2,…,bnb1,b2,…,bn. Find the lexicographically minimal permutation a1,a2,…,a2na1,a2,…,a2n such that bi=min(a2i−1,a2i)bi=min(a2i−1,a2i), or determine that it is impossible.

Input

Each test contains one or more test cases. The first line contains the number of test cases tt (1≤t≤1001≤t≤100).

The first line of each test case consists of one integer nn — the number of elements in the sequence bb (1≤n≤1001≤n≤100).

The second line of each test case consists of nn different integers b1,…,bnb1,…,bn — elements of the sequence bb (1≤bi≤2n1≤bi≤2n).

It is guaranteed that the sum of nn by all test cases doesn't exceed 100100.

Output

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

Otherwise, print 2n2n integers a1,…,a2na1,…,a2n — required lexicographically minimal permutation of numbers from 11 to 2n2n.

Example

input

Copy

output

Copy

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

代码：

#include<bits/stdc++.h>
using namespace std;
long long n,t,r,q,k,s,max1=0,p;
long long a[10001],b[10001];
map<long long ,long long>m;
int main()
{cin>>t;while(t--){cin>>n;m.clear();for(int i=1;i<=n;i++){cin>>b[i];m[b[i]]=1;}int flag=1;for(int i=1;i<=n;i++){a[i*2-1]=b[i];k=b[i];while(m[k]==1){k++;}if(k<=2*n){a[i*2]=k;m[k]=1;}else{flag=0;break;}}if(flag==0){cout<<-1<<endl;}else{for(int i=1;i<=2*n;i++){cout<<a[i]<<" ";}cout<<endl;}}
}

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