1477_D. Nezzar and Hidden Permutations

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.

Example input:
3
4 2
1 2
3 4
6 4
1 2
1 3
3 5
3 6
2 1
1 2


Example 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


Tags: constructive algorithms
dfs and similar
graphs
Difficulty rating: 2800

Solution:
#include<bits/stdc++.h>
using namespace std;
#define debug(x) cerr<<#x<<':'<<x<<endl;
const int N = 5e5+20;
set<int> G[N],S[N];
int id[N],rt[N],p[N],q[N],T,n,m,cnt;
void mk(int x,int y)
{
	rt[id[x]=id[y]=++cnt]=x;
	S[cnt].insert(x);S[cnt].insert(y);
}
void mian()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)G[i].clear(),id[i]=0;
	for(int i=1,x,y;i<=m;i++)
		scanf("%d%d",&x,&y),G[x].insert(y),G[y].insert(x);
	int all=0;cnt=0;
	for(int i=1;i<=n;i++)
	{
		if(G[i].size()==n-1){p[i]=q[i]=++all;continue;}
		if(id[i])continue;
		int to = 1;
		G[i].insert(n+1);
		for(int v:G[i])
		{
			if(i == to)++to;
			if(v!=to)break;
			++to;
		}
		if(!id[to]){mk(i,to);continue;}
		int t = id[to];
		if(S[t].size()==2 || rt[t]==to)
		{
			S[t].insert(i);
			rt[id[i]=t]= to;
		}
		else S[t].erase(to),mk(i,to);
	}
	for(int i=1;i<=cnt;i++)
	{
		p[rt[i]] = ++all;
		for(int j:S[i])
			if(j!=rt[i])q[j]=all,p[j]=++all;
		q[rt[i]] = all;
		S[i].clear();rt[i]=0;
	}
	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");
}
int main()
{
	scanf("%d",&T);
	while(T--)mian();
	return 0;
}