Codeforces 1315C 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

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 题意很简单。因为有要求：1.bi=min(a2i−1,a2i)，2.字典序尽可能

#include <bits/stdc++.h>
using namespace std;
int b[205];
int out[205];
bool vis[205];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        int i,j;
        int flag=1;
        memset(vis,0,sizeof(vis));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&b[i]);
            vis[b[i]]=1;
            out[2*i-1]=b[i];
        }

        for(i=1;i<=n;i++)
        {
            int find=0;

            for(j=1;j<=2*n;j++)
            {
                if(vis[j])continue;
                if(j>out[2*i-1])
                {
                    out[2*i]=j;
                    find=1;
                    vis[j]=1;
                    break;
                }
            }

            if(!find)
            {
                 flag=0;
                 break;
            }
        }

        if(!flag)
        {
            cout<<-1<<endl;
            continue;
        }
        for(i=1;i<=2*n;i++)
        {
            cout<<out[i]<<‘ ‘;
        }
        cout<<endl;
    }
    return 0;
}

小。所以肯定要把bi放到2i-1的位置（观察样例也不难看出）。然后就是在剩下的数里从小到大地找，找到第一个比bi大的数填到a2i的位置，没有这么一个数的话返回-1.至于正确性的话可以这么想，假设b数组里靠前的是比较小的数，那么从剩下的数里从小到大地选，能保证剩下的里较大的数留给后面更大的bi；假设b数组里靠前的是比较大的数，挑过以后肯定还能保证后面较小的bi有数和它搭配，所以贪心是正确的。

原文地址：https://www.cnblogs.com/lipoicyclic/p/12364866.html