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;
namespace Sakurajima_Mai{
    #define ms(a) memset(a,0,sizeof(a))
    #define repi(i,a,b) for(int i=a,bbb=b;i<=bbb;++i)//attention reg int or reg ll ?
    #define repd(i,a,b) for(int i=a,bbb=b;i>=bbb;--i)
    #define reps(s) for(int i=head[s],v=e[i].to;i;i=e[i].nxt,v=e[i].to)
    #define ce(i,r) i==r?'\n':' '
    #define pb push_back 
    #define all(x) x.begin(),x.end()
    #define gmn(a,b) a=min(a,b)
    #define gmx(a,b) a=max(a,b)
    #define fi first
    #define se second                         
    typedef long long ll;
    typedef unsigned long long ull;
    typedef double db;
    const int infi=1e9;//infi较大，注意涉及inf相加时爆int
    const ll infl=4e18;
    inline ll ceil_div(ll a,ll b){ return (a+b-1)/b; }
    inline ll pos_mod(ll a,ll b){ return (a%b+b)%b; }
    //std::mt19937 rnd(time(0));//std::mt19937_64 rnd(time(0));
}
using namespace Sakurajima_Mai;
namespace Fast_Read{
    inline int qi(){
        int f=0,fu=1; char c=getchar();
        while(c<'0'||c>'9'){ if(c=='-')fu=-1; c=getchar(); }
        while(c>='0'&&c<='9'){ f=(f<<3)+(f<<1)+c-48; c=getchar(); }
        return f*fu;
    }
    inline ll ql(){
        ll f=0;int fu=1; char c=getchar();
        while(c<'0'||c>'9'){ if(c=='-')fu=-1; c=getchar(); }
        while(c>='0'&&c<='9'){ f=(f<<3)+(f<<1)+c-48; c=getchar(); }
        return f*fu;
    }
    inline db qd(){
        char c=getchar();int flag=1;double ans=0;
        while((!(c>='0'&&c<='9'))&&c!='-') c=getchar();
        if(c=='-') flag=-1,c=getchar();
        while(c>='0'&&c<='9') ans=ans*10+(c^48),c=getchar();
        if(c=='.'){c=getchar();for(int Bit=10;c>='0'&&c<='9';Bit=(Bit<<3)+(Bit<<1)) ans+=(double)(c^48)*1.0/Bit,c=getchar();}
        return ans*flag;
    }
}
namespace Read{
    #define si(a) scanf("%d",&a)
    #define sl(a) scanf("%lld",&a)
    #define sd(a) scanf("%lf",&a)
    #define ss(a) scanf("%s",a)
    #define rai(x,a,b) repi(i,a,b) x[i]=qi()
    #define ral(x,a,b) repi(i,a,b) x[i]=ql()
}
namespace Out{
    #define pi(x) printf("%d",x)
    #define pl(x) printf("%lld",x)
    #define ps(x) printf("%s",x)
    #define pc(x) printf("%c",x)
    #define pe() puts("")
}
namespace DeBug{
    #define MARK false
    #define DB if(MARK) 
    #define pr(x) cout<<#x<<": "<<x<<endl
    #define pra(x,a,b) cout<<#x<<": "<<endl; \
            repi(i,a,b) cout<<x[i]<<" ";     \
            puts("");   
    #define FR(a) freopen(a,"r",stdin)
    #define FO(a) freopen(a,"w",stdout)
}
using namespace Fast_Read;
using namespace Read;
using namespace Out;
using namespace DeBug;
const int MAX_N=2e5+5;
int n,q[MAX_N];
int ans[MAX_N];
set<int> s;
void solve_min()
{
	repi(i,1,n) s.insert(i);
	ans[1]=q[1],s.erase(q[1]);
	repi(i,2,n){
		if(q[i]==q[i-1]) ans[i]=*s.begin();
		else ans[i]=q[i];
		s.erase(ans[i]);
	}
	repi(i,1,n) pi(ans[i]),pc(' ');
	pe();
}
void solve_max()
{
	ans[1]=q[1];
	repi(i,1,q[1]-1) s.insert(i);
	int last=q[1];
	repi(i,2,n){
		if(q[i]==q[i-1]) ans[i]=*s.rbegin(),s.erase(ans[i]);
		else{
			ans[i]=q[i];
			repd(j,q[i]-1,last+1) s.insert(j);
			last=q[i];
		}
	}
	repi(i,1,n) pi(ans[i]),pc(' ');
	pe();
}
int main()
{
	int T=qi();
	while(T--)
	{
		n=qi();
		rai(q,1,n);
		solve_min(),solve_max();
	}
    return 0;
}