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<bits/stdc++.h>
using namespace std;
const int MAXN = 5e5 + 5;
set<int> G[MAXN];
set<int> F[MAXN];
set<int> V;
int p1[MAXN], p2[MAXN];
struct star {
	int rt;
	vector<int> leaves;
};
vector<star> stars;
set<pair<int, int> > d;
int n, m, t;
int deg[MAXN];
void init() {
	for(int i = 1; i <= n; i++) {
		G[i].clear();
		F[i].clear();
	}
	V.clear();
	stars.clear();
	d.clear();
}
void dfs(int u) {	
	V.erase(u);
	int crt = 0;
	while(true) {
		auto iter = V.upper_bound(crt);
		if(iter == V.end()) break;
		int v = crt = (*iter);
		if(G[u].find(v) != G[u].end()) continue;
		F[u].insert(v), F[v].insert(u);
		dfs(v);
	}
}
int main() {
	scanf("%d", &t);
	while(t--) {
		scanf("%d%d", &n, &m);
		init();
		for(int i = 1; i <= m; i++) {
			int l, r;
			scanf("%d%d", &l, &r);
			G[l].insert(r); G[r].insert(l);
		}
		for(int i = 1; i <= n; i++) V.insert(i);
		while(V.size()) {
			dfs(*V.begin());	
		}
		int rem = n;
		for(int i = 1; i <= n; i++) {
			deg[i] = F[i].size();	
			if(deg[i]) {
				d.insert(make_pair(deg[i], i));
			} else {
				p1[i] = p2[i] = rem;
				rem--;
			}
		}
		while(d.size()) {
			int id = (*d.begin()).second;
			int u = (*F[id].begin()); 
			vector<int> leaves;
			leaves.clear();
			d.erase(make_pair(deg[u], u));
			for(int v : F[u]) {
				d.erase(make_pair(deg[v], v));
				if(deg[v] == 1) {
					leaves.push_back(v);
				} else {
					deg[v]--;
					d.insert(make_pair(deg[v], v));
					F[v].erase(u);
				}
			}
			stars.push_back((star){ u, leaves});
		}
		int st = 0, en = 0;
		for(auto i : stars) {
			p1[i.rt] = ++st;
			for(int j : i.leaves) {
				p1[j] = ++st;
				p2[j] = ++en;
			}
			p2[i.rt] = ++en;
		}
		for(int i = 1; i <= n; i++) printf("%d ", p1[i]);
		printf("\n");
		for(int i = 1; i <= n; i++) printf("%d ", p2[i]);
		printf("\n");
	}
	return 0;
}
	 			  		     	  	 	  						 		