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 <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__);fflush(stderr);
#else
	#define eprintf(...) 42
#endif

using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
template<typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll myRand(ll B) {
	return (ull)rng() % B;
}

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

clock_t startTime;
double getCurrentTime() {
	return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}

const int N = 500500;
int n, m;
vector<int> a[N];
vector<int> b[N];
int myId[N];
vector<int> g[N];
int ans[2][N];
bool used[N];

void addAns(vector<int> z) {
	for (int v : z)
		myId[v] = m;
	b[m++] = z;
}

void dfs(int v) {
	used[v] = 1;
	vector<int> tt;
	for (int u : g[v]) {
		if (used[u]) continue;
		dfs(u);
		if (myId[u] == -1)
			tt.push_back(u);
	}
	if (!tt.empty()) {
		tt.push_back(v);
		reverse(all(tt));
		addAns(tt);
	}
}

void solve() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++) {
		a[i].clear();
		b[i].clear();
		g[i].clear();
		myId[i] = -1;
		used[i] = 0;
	}
	while(m--) {
		int v, u;
		scanf("%d%d", &v, &u);
		v--;u--;
		a[v].push_back(u);
		a[u].push_back(v);
	}
	m = 0;
	for (int v = 0; v < n; v++) {
		sort(all(a[v]));
		a[v].resize(unique(a[v].begin(), a[v].end()) - a[v].begin());
		if ((int)a[v].size() == n - 1) continue;
		int u = 0;
		for (int x : a[v]) {
			if (u == v) u++;
			if (u == x) {
				u++;
			} else {
				break;
			}
		}
		if (u == v) u++;
		g[v].push_back(u);
		g[u].push_back(v);
	}
	for (int v = 0; v < n; v++) {
		if (used[v] || g[v].empty()) continue;
		dfs(v);
		if (myId[v] == -1) {
			assert(m > 0);
			b[m - 1].push_back(v);
			myId[v] = m - 1;
		}
	}
	int k = 1;
	for (int v = 0; v < n; v++) {
		if (myId[v] != -1) continue;
		assert(g[v].empty());
		ans[0][v] = ans[1][v] = k++;
	}
	for (int i = 0; i < m; i++) {
		for (int j = 1; j < (int)b[i].size(); j++) {
			int v = b[i][j];
			ans[0][v] = k + j - 1;
			ans[1][v] = k + j;
		}
		int v = b[i][0];
		ans[0][v] = k + (int)b[i].size() - 1;
		ans[1][v] = k;
		k += (int)b[i].size();
	}
	for (int i = 0; i < 2; i++) {
		for (int j = 0; j < n; j++)
			printf("%d ", ans[i][j]);
		printf("\n");
	}
}

int main()
{
	startTime = clock();
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	int t;
	scanf("%d", &t);
	while(t--) solve();

	return 0;
}
