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:
/***************************************************************
	File name: CF1477D.cpp
	Author: ljfcnyali
	Create time: 2021年01月29日 星期五 11时14分42秒
***************************************************************/
#include<bits/stdc++.h>

using namespace std;

#define REP(i, a, b) for ( int i = (a), _end_ = (b); i <= _end_; ++ i ) 
#define mem(a) memset ( (a), 0, sizeof ( a ) ) 
#define str(a) strlen ( a ) 
#define pii pair<int, int>
typedef long long LL;

const int maxn = 1e6 + 10;

int n, m, T,  a[maxn], b[maxn], Begin[maxn], Next[maxn], To[maxn], e, tot, id[maxn];
vector<int> Edge[maxn], Set[maxn];
bool vis[maxn];

inline void add(int u, int v) { To[++ e] = v; Next[e] = Begin[u]; Begin[u] = e; } 

inline bool DFS(int u)
{
	vis[u] = true;
	for ( auto v : Edge[u] ) if ( !vis[v] && !DFS(v) )
	{
		if ( !id[u] ) { id[u] = ++ tot; Set[id[u]].push_back(u); }
		Set[id[u]].push_back(v);
	}
	if ( !id[u] ) return false;
	return true;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
	scanf("%d", &T);
	while ( T -- ) 
	{
		scanf("%d%d", &n, &m);
		e = tot = 0; REP(i, 1, n) { Edge[i].clear(); Set[i].clear(); Begin[i] = vis[i] = id[i] = 0; } 
		REP(i, 1, m) { int u, v; scanf("%d%d", &u, &v); add(u, v); add(v, u); } 
		REP(i, 1, n)	
		{
			for ( int j = Begin[i]; j; j = Next[j] ) vis[To[j]] = true;
			int pos = -1; REP(j, 1, n) if ( i != j && !vis[j] ) { pos = j; break; } 
			if ( pos == -1 ) Set[++ tot].push_back(i);
			else { Edge[i].push_back(pos); Edge[pos].push_back(i); }
			for ( int j = Begin[i]; j; j = Next[j] ) vis[To[j]] = false;
		}
		REP(i, 1, n) if ( !vis[i] && Edge[i].size() ) if ( !DFS(i) ) Set[id[Edge[i][0]]].push_back(i);

		int now = 0;
		REP(i, 1, tot) 
		{
			a[Set[i][0]] = now + 1; b[Set[i][0]] = now + Set[i].size();
			for ( int j = 1; j < Set[i].size(); ++ j ) a[Set[i][j]] = now + j + 1, b[Set[i][j]] = now + j;
			now += Set[i].size();
		}

		REP(i, 1, n) printf("%d ", a[i]); puts("");
		REP(i, 1, n) printf("%d ", b[i]); puts("");
		REP(u, 1, n) for ( int i = Begin[u]; i; i = Next[i] ) { int v = To[i]; assert(((a[u] <= a[v]) ^ (b[u] <= b[v])) == 0); }
	}
    return 0;
}