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 <algorithm>
#include <vector>
using namespace std;
//#define endl '\n'
#define ll long long
#define pi pair<int, int>
#define f first
#define s second

const int mxn = 500000;
int n, m, k;
int a[mxn], p[mxn], q[mxn], vis[mxn];
vector<int> g[mxn], gr[mxn], v[mxn];

bool dfs(int c){
	vis[c] = 1;
	for(int i : gr[c]) if(!vis[i]){
		if(dfs(i)){
			if(!~a[c]) v[a[c] = k++].push_back(c);
			v[a[i] = a[c]].push_back(i);
		}
	}
	return !~a[c];
}

void answer(){
	cin >> n >> m;
	k = 0;
	
	for(int i = 0; i < n; i++){
		a[i] = -1, vis[i] = 0;
		g[i].assign(1, i), gr[i].clear(), v[i].clear();
	} 
	
	for(int i = 0; i < m; i++){
		int x, y;
		cin >> x >> y;
		x--, y--;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	
	for(int i = 0; i < n; i++){
		int x = 0;
		for(int j : g[i]) vis[j] = 1;
		while(x < n && vis[x]) x++;
		if(x < n) gr[i].push_back(x), gr[x].push_back(i);
		else v[a[i] = k++].push_back(i);
		for(int j : g[i]) vis[j] = 0;
	}

	for(int i = 0; i < n; i++) if(!gr[i].empty() && !vis[i]){
		if(dfs(i)) a[i] = a[gr[i][0]], v[a[i]].push_back(i);
	}

	for(int i = 0, x = 0; i < k; i++){
		for(int j = 0; j < v[i].size(); j++){
			p[v[i][j]] = x + j + 1, q[v[i][j]] = x + j;
		} 
		x += v[i].size(), q[v[i][0]] = x;
	}
	
	cout << p[0];
	for(int i = 1; i < n; i++) cout << " " << p[i];
	cout << endl;
	cout << q[0];
	for(int i = 1; i < n; i++) cout << " " << q[i];
	cout << endl;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	int t;
	cin >> t;
	
	for(int i = 0; i < t; i++) answer();

	return 0;
}