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.
using namespace std;
#include <bits/stdc++.h>
#define N 500005
int n,m;
set<int> e[N];
int d[N];
int p[N],q[N];
int to[N];
int bel[N];
vector<int> vec[N];
int main(){
//	freopen("in.txt","r",stdin);
	int T;
	scanf("%d",&T);
	while (T--){
		scanf("%d%d",&n,&m);
		for (int i=1;i<=n;++i)
			e[i].clear();
		memset(d,0,sizeof(int)*(n+1));
		for (int i=1;i<=m;++i){
			int u,v;
			scanf("%d%d",&u,&v);
			e[u].insert(v);
			e[v].insert(u);
			d[u]++,d[v]++;
		}
		memset(p,0,sizeof(int)*(n+1));
		memset(q,0,sizeof(int)*(n+1));
		for (int i=1;i<=n;++i)
			if (d[i]==n-1)
				p[i]=q[i]=-1;
			else{
				for (int j=1;j<=n;++j)
					if (j!=i && e[i].find(j)==e[i].end()){
						to[i]=j;
						break;
					}
			}
		for (int i=1;i<=n;++i)
			vec[i].clear();
		memset(bel,0,sizeof(int)*(n+1));
		for (int i=1;i<=n;++i){
			if (d[i]==n-1) continue;
			if (!bel[i] && !bel[to[i]])
				bel[i]=to[i],bel[to[i]]=i;
		}
		for (int i=1;i<=n;++i)
			if (d[i]!=n-1 && bel[i]==0)
				vec[to[i]].push_back(i);
		int c=0;
		for (int i=1;i<=n;++i){
			if (d[i]==n-1) continue;
			if (bel[i] && i<bel[i]){
				int x=i,y=bel[i];
				if (vec[x].empty())
					swap(x,y);
				if (vec[x].empty()){
					p[x]=c+1,q[x]=c+2;
					p[y]=c+2,q[y]=c+1;
					c+=2;
				}
				else{
					if (vec[y].empty()){
						int t=c;
						++t,p[y]=t,q[y]=t+1;
						for (int j=0;j<vec[x].size();++j)
							++t,p[vec[x][j]]=t,q[vec[x][j]]=t+1;
						++t,p[x]=t,q[x]=c+1;
						c=t;
					}
					else{
						int t=c;
						for (int j=0;j<vec[x].size();++j)
							++t,p[vec[x][j]]=t,q[vec[x][j]]=t+1;
						++t,p[x]=t,q[x]=c+1;
						c=t;
						for (int j=0;j<vec[y].size();++j)
							++t,p[vec[y][j]]=t,q[vec[y][j]]=t+1;
						++t,p[y]=t,q[y]=c+1;
						c=t;
					}
				}
			}
		}
		for (int i=1;i<=n;++i)
			if (d[i]==n-1)
				p[i]=q[i]=++c;
		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");
	}
	return 0;
}
















