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<iostream>
#include<cstdio>
#include<set>
#include<vector>
#include<cstring>
#define N 500010
using namespace std;
set<int>s,v[N];
vector<int>tr[N],num[N];
int vis[N],de[N],color,size[N],col[N],root[N],pd[N],T,n,m,now1,now2,p[N],q[N];
void add_edge(int x,int y){
	v[x].insert(y);
	v[y].insert(x);
}
void ADD(int x,int y){
	tr[x].push_back(y);
	tr[y].push_back(x);
}
void dfs1(int x,int f){
//	cout<<"check "<<x<<" "<<f<<endl;
	vis[x]=1;
	set<int>::iterator it;
	for(it=s.begin();it!=s.end();it++)if(!v[x].count(*it))ADD(x,*it);
	for(int i=0;i<tr[x].size();i++){
		int t=tr[x][i];
		if(t==f)continue;
		s.erase(t);
	}
	for(int i=0;i<tr[x].size();i++){
		int t=tr[x][i];
		if(t==f)continue;
		dfs1(t,x);
	}
}
void dfs2(int x,int f){
//	cout<<"check "<<x<<' '<<f<<endl;
	if(!col[x]){
		int flag=0;
		for(int i=0;i<tr[x].size();i++){
			int t=tr[x][i];
			if(t==f)continue;
			if(!col[t]){
				flag=1;
				break;
			}
		}
		if(flag){
			color++;
			root[color]=x;
			size[color]=1;
			col[x]=color;
			for(int i=0;i<tr[x].size();i++){
				int t=tr[x][i];
				if(t==f)continue;
				if(!col[t])col[t]=color,size[color]++;
			}
		}
		else{
			for(int i=0;i<tr[x].size();i++){
				int t=tr[x][i];
				if(t==f)continue;
				if(col[t]){
					if(root[col[t]]==t){
						col[x]=col[t];
						size[col[t]]++;
					}
					else if(size[col[t]]>2){
						color++;
						size[col[t]]--;
						root[color]=x;
						col[x]=col[t]=color;
						size[color]=2;
					}
					else{
						root[col[t]]=t;
						size[col[t]]++;
						col[x]=col[t];
					}
				}
				break;
			}
		}
	}
	for(int i=0;i<tr[x].size();i++){
		int t=tr[x][i];
		if(t==f)continue;
		dfs2(t,x);
	}
}
int main(){
	cin>>T;
	while(T--){
		cin>>n>>m;
		for(int i=1;i<=n;i++){
			while(v[i].size())v[i].erase(v[i].begin());
			while(num[i].size())num[i].pop_back();
			while(tr[i].size())tr[i].pop_back();
		}
		for(int i=1;i<=n;i++)s.insert(i);
		for(int i=1;i<=n;i++){
			size[i]=vis[i]=pd[i]=root[i]=de[i]=col[i]=0;
		}
		color=now1=now2=0;
		for(int i=1;i<=m;i++){
			int x,y;
			scanf("%d%d",&x,&y);
			add_edge(x,y);
			add_edge(y,x);
			de[x]++,de[y]++;
		}
		for(int i=1;i<=n;i++){
			if(de[i]==n-1){
				s.erase(i),pd[i]=1;
				now1++,now2++;
				p[i]=q[i]=now1;
			}
		}
		for(int i=1;i<=n;i++)if(!vis[i]&&!pd[i])dfs1(i,0),s.erase(i);
		for(int i=1;i<=n;i++)if(!col[i]&&!pd[i])dfs2(i,0);
		for(int i=1;i<=n;i++){
			if(root[col[i]]!=i)num[col[i]].push_back(i);
		}
		for(int i=1;i<=color;i++){
			now1++;
			p[root[i]]=now1;
			for(int j=0;j<num[i].size();j++){
				p[num[i][j]]=++now1;
				q[num[i][j]]=++now2;
			}
			now2++;
			q[root[i]]=now2;
		}
		for(int i=1;i<=n;i++)printf("%d ",p[i]);
		printf("\n");
		for(int i=1;i<=n;i++)printf("%d ",q[i]);
		printf("\n");
	}
	
}//xwp
		 	 	 	 	  	    	 		  	 		   	