Nezzar designs a brand new game "Hidden Permutations" and shares it with his best friend, Nanako.

At the beginning of the game, Nanako and Nezzar both know integers n and m. The game goes in the following way:

  * Firstly, Nezzar hides two permutations p_1,p_2,…,p_n and q_1,q_2,…,q_n of integers from 1 to n, and Nanako secretly selects m unordered pairs (l_1,r_1),(l_2,r_2),…,(l_m,r_m); 
  * After that, Nanako sends his chosen pairs to Nezzar; 
  * On receiving those m unordered pairs, Nezzar checks if there exists 1 ≤ i ≤ m, such that (p_{l_i}-p_{r_i}) and (q_{l_i}-q_{r_i}) have different signs. If so, Nezzar instantly loses the game and gets a score of -1. Otherwise, the score Nezzar gets is equal to the number of indices 1 ≤ i ≤ n such that p_i ≠ q_i. 



However, Nezzar accidentally knows Nanako's unordered pairs and decides to take advantage of them. Please help Nezzar find out two permutations p and q such that the score is maximized.

Input

The first line contains a single integer t (1 ≤ t ≤ 5 ⋅ 10^5) — the number of test cases.

The first line of each test case contains two integers n,m (1 ≤ n ≤ 5 ⋅ 10^5, 0 ≤ m ≤ min((n(n-1))/(2),5 ⋅ 10^5)). 

Then m lines follow, i-th of them contains two integers l_i,r_i (1 ≤ l_i,r_i ≤ n, l_i ≠ r_i), describing the i-th unordered pair Nanako chooses. It is guaranteed that all m unordered pairs are distinct.

It is guaranteed that the sum of n for all test cases does not exceed 5 ⋅ 10^5, and the sum of m for all test cases does not exceed 5⋅ 10^5.

Output

For each test case, print two permutations p_1,p_2,…,p_n and q_1,q_2,…,q_n such that the score Nezzar gets is maximized.

Example

Input


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


Output


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

Note

For first test case, for each pair given by Nanako:

  * for the first pair (1,2): p_1 - p_2 = 1 - 2 = -1, q_1 - q_2 = 3 - 4 = -1, they have the same sign; 
  * for the second pair (3,4): p_3 - p_4 = 3 - 4 = -1, q_3 - q_4 = 1 - 2 = -1, they have the same sign. 



As Nezzar does not lose instantly, Nezzar gains the score of 4 as p_i ≠ q_i for all 1 ≤ i ≤ 4. Obviously, it is the maximum possible score Nezzar can get.
#include<bits/stdc++.h>
#define For(i,a,b) for(register int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(register int i=(a);i>=(b);--i)
using namespace std;
inline int read()
{
    char c=getchar();int x=0;bool f=0;
    for(;!isdigit(c);c=getchar())f^=!(c^45);
    for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
    if(f)x=-x;return x;
}
#define maxn 500005
#define pb push_back
int n,m,du[maxn],cnt,res[maxn];
bool ct[maxn];
vector<int>e[maxn],cen[maxn];
set<int>G[maxn],vis;
inline void adde(int u,int v){e[u].pb(v),e[v].pb(u);}
vector<int>p,q;
void build(int rt,vector<int>u){
	if(!u.size())exit(233);
	p.pb(rt);
	for(auto x:u)p.pb(x),q.pb(x);
	q.pb(rt);
}
void print(vector<int>p){
	if(p.size()!=n)exit(2333);
	For(i,1,n)res[p[i-1]]=i;
	For(i,1,n)printf("%d ",res[i]);puts("");
}

void bfs(int u)
{
	queue<int>q; q.push(u),vis.erase(u);
	while(!q.empty()){
		int u=q.front();q.pop();
		vector<int>adj;
		for(auto v:vis)if(!G[u].count(v))adj.pb(v);
		for(auto v:adj)vis.erase(v),q.push(v),adde(u,v);
	}
}

void dfs(int u,int pa)
{
	for(auto v:e[u])
		if(v!=pa){
			dfs(v,u);
			if(ct[u]&&!ct[v])cen[u].pb(v);
		}
	if(!ct[u]){
		if(pa)ct[pa]=1;
		else{
			for(auto v:e[u])
				if(ct[v]){cen[v].pb(u);break;}
		}
	}
}

void work()
{
	n=read(),m=read();p.clear(),q.clear(),vis.clear();
	For(u,1,n)G[u].clear(),e[u].clear(),cen[u].clear(),du[u]=ct[u]=0;
	For(u,1,n)vis.insert(u);
	For(i,1,m){
		int u=read(),v=read();
		G[u].insert(v),G[v].insert(u),du[u]++,du[v]++;
	}
	cnt=0;
	For(i,1,n)if(du[i]==n-1)++cnt,vis.erase(i),p.pb(i),q.pb(i);
	For(i,1,n)if(vis.count(i))bfs(i),dfs(i,0);
	For(i,1,n)if(ct[i])build(i,cen[i]);
	print(p),print(q);
}

signed main()
{
	int T=read();
	while(T--)work();
    return 0;
}
/*
3
4 2
1 2
3 4
*/