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 <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 500005
int n, m, x, y, tot, A[N], B[N], cnt, idx[N];
vector<int> G[N], E[N], P[N];
bool vis[N];
bool Dfs(int x)
{
	vis[x]=1;
	for (int i = 0; i < (int)E[x].size(); i++)
		if (!vis[E[x][i]])
			if (!Dfs(E[x][i]))
			{
				if (!idx[x])
				{
					idx[x]=++tot;
					P[tot].push_back(x);
				}
				P[idx[x]].push_back(E[x][i]);
			}
	if (!idx[x])
		return 0;
	return 1;
}
int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d", &n, &m);
		tot = 0;
		for (int i = 1; i <= n; i++)
		{
			G[i].clear();
			E[i].clear();
			vis[i] = 0, idx[i] = 0;
		}
		for (int i = 1; i <= m; i++)
		{
			scanf("%d%d", &x, &y);
			G[x].push_back(y);
			G[y].push_back(x);
		}
		for (int i = 1; i <= n; i++)
		{
			int p = 0;
			for (int j = 0; j < (int)G[i].size(); j++)
				vis[G[i][j]] = 1;
			for (int j = 1; j <= n; j++)
				if (i != j && !vis[j])
				{
					p = j;
					break;
			}
			if (!p)
				P[++tot].push_back(i);
			else
			{
				E[i].push_back(p);
				E[p].push_back(i);
			}
			for (int j = 0; j < (int)G[i].size(); j++)
				vis[G[i][j]] = 0;
		}
		for (int i = 1; i <= n; i++)
			if (!vis[i] && E[i].size())
				if (!Dfs(i))
					P[idx[E[i][0]]].push_back(i);
		cnt = 0;
		for (int i = 1; i <= tot; i++)
		{
			int sz = P[i].size();
			for (int j = 0; j<sz; j++)
			{
				A[P[i][j]] = cnt + j + 1;
				B[P[i][j]] = cnt + j;
			}
			cnt += sz;
			B[P[i][0]] = cnt;
			P[i].clear();
		}
		for (int i = 1; i <= n; i++)
			printf("%d%c", A[i], i == n ? 10 : 32);
		for (int i = 1; i <= n; i++)
			printf("%d%c", B[i], i == n ? 10 : 32);;
	}
}

