Codeforces Round #623 C. Restoring Permutation

C. Restoring Permutation

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

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

output

Copy

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

题意 :  给一组 长度为 n 的  b[i] , 让你找一组 长度 为 2n 的a[i] , 满足 b[i] = min(a[2*i-1 ]  ,a[2*i] ) ,并且字典序最小 。

思路 : 贪心 

其实就是在左边下面的图中填空,右边空白的要大于左边的数字,并且要先从 左边开始模拟我们填数的过程。

#include <iostream> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
using namespace std ;
typedef long long LL  ; 
/* 

*/
const int MAX = 51000 ; 
const int inf = 0x3f3f3f3f ;  
/*
 给 b1 ,b2 .. bn ; 
 找字典序最小的 a1 , ... a2n 
 bi = min(a_2i-1  ,a_2i)


*/
int vis[MAX] ; 
int b[MAX] ; 
int a[MAX] ; 
int main(){
    int t ; 
    cin >> t; 
    while(t--){
        int n ; 
        cin >>n ; 
        memset(vis,0,sizeof(vis));
        for(int i = 1 ;i<=n ; i++ ) {
            cin >>b[i] ; 
            a[2*i-1] = b[i] ;
            vis[b[i]] = 1 ; 
        }
        bool flag = true ; 
        for(int i = 1 ; i<=n ;i++ ) {
            int t = b[i] ; 
            while(vis[t]){
                t++ ; 
            }
            
            if(t<=2*n) {
                a[2*i] = t ;
                vis[t] = 1 ;
            } 
            else {
                flag = false ; 
                break ; 
            }
             
        }
        // bool flag = true ; 
        // for(int i = 1 ;i<=2*n ;i++) {
        //     if(b[(i+1 )/2] > a[i]){
        //         flag = false ; 
        //         break ; 
        //     }
        // }
        if(flag ) {
            for(int i = 1 ; i<=2*n ; i++) {
                printf("%d ",a[i]);
            }
        }
        else{
            printf("-1") ; 
        }


        printf("\n") ; 
        
    }

    
    return 0 ; 
}