Problem: A permutation is a sequence of n integers from 1 to n, in which all numbers occur exactly once. For example, [1], [3, 5, 2, 1, 4], [1, 3, 2] are permutations, and [2, 3, 2], [4, 3, 1], [0] are not.

Polycarp was presented with a permutation p of numbers from 1 to n. However, when Polycarp came home, he noticed that in his pocket, the permutation p had turned into an array q according to the following rule: 

  * q_i = max(p_1, p_2, …, p_i). 



Now Polycarp wondered what lexicographically minimal and lexicographically maximal permutations could be presented to him.

An array a of length n is lexicographically smaller than an array b of length n if there is an index i (1 ≤ i ≤ n) such that the first i-1 elements of arrays a and b are the same, and the i-th element of the array a is less than the i-th element of the array b. For example, the array a=[1, 3, 2, 3] is lexicographically smaller than the array b=[1, 3, 4, 2].

For example, if n=7 and p=[3, 2, 4, 1, 7, 5, 6], then q=[3, 3, 4, 4, 7, 7, 7] and the following permutations could have been as p initially: 

  * [3, 1, 4, 2, 7, 5, 6] (lexicographically minimal permutation); 
  * [3, 1, 4, 2, 7, 6, 5]; 
  * [3, 2, 4, 1, 7, 5, 6]; 
  * [3, 2, 4, 1, 7, 6, 5] (lexicographically maximum permutation). 



For a given array q, find the lexicographically minimal and lexicographically maximal permutations that could have been originally presented to Polycarp.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).

The second line of each test case contains n integers q_1, q_2, …, q_n (1 ≤ q_i ≤ n).

It is guaranteed that the array q was obtained by applying the rule from the statement to some permutation p.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output two lines: 

  * on the first line output n integers — lexicographically minimal permutation that could have been originally presented to Polycarp; 
  * on the second line print n integers — lexicographically maximal permutation that could have been originally presented to Polycarp; 

Example

Input


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


Output


3 1 4 2 7 5 6 
3 2 4 1 7 6 5 
1 2 3 4 
1 2 3 4 
3 4 5 1 2 7 6 
3 4 5 2 1 7 6 
1 
1 

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define maxn 200005
#define maxm 200005
#define INF 1234567890
#define eps 1e-15
#define p 1000000007
const double pi=acos(-1);
template<class T>inline bool read(T &x)
{
    x=0;register char c=getchar();register bool f=0;
    while(!isdigit(c)){if(c==EOF)return false;f^=c=='-',c=getchar();}
    while(isdigit(c))x=(x<<3)+(x<<1)+(c^48),c=getchar();
    if(f)x=-x;
    return true;
}
template<class T>inline void print(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x>9)print(x/10);
    putchar(x%10^48);
}
template<class T>inline void print(T x,char c){print(x),putchar(c);}
template<class T,class ...S>inline bool read(T &x,S &...y){return read(x)&&read(y...);}
ll T,n,a[maxn],b[maxn],x[maxn],mx,nxt[maxn],pre[maxn],l;
bool vis[maxn];
int main()
{
    //freopen("test.in","r",stdin);
    read(T);
    while(T--)
    {
        read(n);
        for(ll i=1;i<=n;i++)
        {
            read(x[i]);
            vis[i]=0;
        }
        for(ll i=0;i<=n;i++)nxt[i]=i+1;
        for(ll i=1;i<=n+1;i++)pre[i]=i-1;
        mx=0;
        l=1;
        for(ll i=1;i<=n;i++)
        {
            if(x[i]>mx)
            {
                a[i]=x[i];
                mx=max(mx,x[i]);
                vis[x[i]]=1;
            }
            else
            {
                for(ll j=l;j<=n;j++)
                {
                    if(!vis[j])
                    {
                        a[i]=j;
                        l=j+1;
                        break;
                    }
                }
            }
        }
        for(ll i=0;i<=n;i++)nxt[i]=i+1;
        for(ll i=1;i<=n+1;i++)pre[i]=i-1;
        mx=0;
        for(ll i=1;i<=n;i++)
        {
            if(x[i]>mx)
            {
                b[i]=x[i];
                mx=max(mx,x[i]);
                nxt[pre[x[i]]]=nxt[x[i]];
                pre[nxt[x[i]]]=pre[x[i]];
            }
            else
            {
                b[i]=pre[x[i]];
                nxt[pre[pre[x[i]]]]=nxt[pre[x[i]]];
                pre[nxt[pre[x[i]]]]=pre[pre[x[i]]];
                pre[x[i]]=pre[pre[x[i]]];
            }
        }
        for(ll i=1;i<=n;i++)print(a[i]," \n"[i==n]);
        for(ll i=1;i<=n;i++)print(b[i]," \n"[i==n]);
    }
    return 0;
}