Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

void solve() {
	int n, m;
	cin >> n >> m;
	set <int> g[n], g1[n];
	for (int i = 0; i < m; ++i) {
		int v, u;
		cin >> v >> u;
		--v;
		--u;
		g[v].insert(u);
		g[u].insert(v);
		g1[v].insert(u);
		g1[u].insert(v);
	}
	vector <int> bed;
	vector <int> used(n);
	for (int i = 0; i < n; ++i) {
		if (g[i].size() + bed.size() == n - 1) {
			bed.push_back(i);
			used[i] = 1;
			for (auto u : g[i]) {
				g[u].erase(i);
			}
		}
	}
	vector <vector <int> > guds;
	set <pair <int, int> > q;
	set <int> vers;
	for (int i = 0; i < n; ++i) {
		if (!used[i]) {
			q.insert({g[i].size(), i});
			vers.insert(i);
		}
	}
	while (!q.empty()) {
		int v = q.rbegin()->second;
		q.erase({g[v].size(), v});
		int u;
		for (auto el : vers) {
			if (el != v && g[v].find(el) == g[v].end()) {
				u = el;
				break;
			}
		}
		q.erase({g[u].size(), u});
		for (auto el : g[v]) {
			q.erase({g[el].size(), el});
			g[el].erase(v);
			q.insert({g[el].size(), el});
		}
		for (auto el : g[u]) {
			q.erase({g[el].size(), el});
			g[el].erase(u);
			q.insert({g[el].size(), el});
		}
		vers.erase(v);
		vers.erase(u);
		vector <int> kek = {u, v};
		while (!q.empty() && q.rbegin()->first == vers.size() - 1) {
			int nxt = q.rbegin()->second;
			kek.push_back(nxt);
			q.erase({g[nxt].size(), nxt});
			vers.erase(nxt);
			if (g1[nxt].find(v) == g1[nxt].end()) swap(kek[0], kek[1]);
			for (auto el : g[nxt]) {
				q.erase({g[el].size(), el});
				g[el].erase(nxt);
				q.insert({g[el].size(), el});
			}
		}
		guds.push_back(kek);
	}
	vector <int> p(n), Q(n);
	for (int i = 1; i <= n; ++i) {
		if (!bed.empty()) {
			p[bed.back()] = i;
			Q[bed.back()] = i;
			bed.pop_back();
			continue;
		}
		auto kek = guds.back();
		p[kek[0]] = i;
		Q[kek[0]] = i + kek.size() - 1;
		for (int j = 1; j < kek.size(); ++j) {
			p[kek[j]] = i + j;
			Q[kek[j]] = i + j - 1;
		}
		i += kek.size() - 1;
		guds.pop_back();
	}
	for (auto el : p) cout << el << ' ';
	cout << '\n';
	for (auto el : Q) cout << el << ' ';
	cout << '\n';
}

signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin >> t;
	while (t) {
		--t;
		solve();
	}
	return 0;
}