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 <vector>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
using namespace std;

#define ll long long
const int MX = 5000005;

int n, m;
set<int> adj[MX];
set<int> g[MX]; // g = dfs tree

struct stars{
	int rt;
	vector<int> leaves;
};

int main(){
	cin.tie(0) -> sync_with_stdio(0);

	int tc; cin >> tc;
	for(; tc --;){
		cin >> n >> m;

		for(int i = 1; i <= n; i ++)
			adj[i].clear(), g[i].clear();

		for(int u, v, i = 0; i < m; i ++){
			cin >> u >> v;
			adj[u].insert(v);
			adj[v].insert(u);
		}

		set<int> vv;
		for(int i = 1; i <= n; i ++)
			vv.insert(i);

		function <void(int)> dfs = [&](int nw){
			vv.erase(nw);
			int nx = 0;

			for(;;){
				auto it = vv.upper_bound(nx);
				if(it == vv.end()) break;

				nx = *it;
				if(adj[nw].count(nx)) continue;

				g[nw].insert(nx);
				g[nx].insert(nw);
				dfs(nx);
			}
		};

		for(; vv.size(); dfs(*vv.begin()));
		
		set<pair<int, int> > ver;
		int cnt = n;
		vector<int> ansA(n + 1), ansB(n + 1);

		for(int i = 1; i <= n; i ++){
			if(g[i].size()) ver.insert({g[i].size(), i});
			else{
				ansA[i] = ansB[i] = cnt;
				cnt --;
			}
		}

		vector<stars> x;
		while(ver.size()){
			int id = (*ver.begin()).second;
			int nx = *g[id].begin();
			ver.erase({g[nx].size(), nx});
			vector<int> leaves;

			for(int around : g[nx]){
				ver.erase({g[around].size(), around});
				if(g[around].size() == 1)
					leaves.push_back(around);
				else{
					g[around].erase(nx);
					ver.insert({g[around].size(), around});
				}
			}

			x.push_back((stars){nx, leaves});
		}

		int lf = 1, rg = 1;
		for(auto i : x){
			ansA[i.rt] = lf ++;
			for(int nx : i.leaves){
				ansA[nx] = lf ++;
				ansB[nx] = rg ++;
			}
			ansB[i.rt] = rg ++;
		}

		for(int i = 1; i <= n; i ++)
			cout << ansA[i] << " ";
		cout << endl;
		for(int i = 1; i <= n; i ++)
			cout << ansB[i] << " ";
		cout << endl;
	}
}