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 <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
using namespace std;
#define ll long long
#define ull unsigned long long
#define fi first
#define se second
#define dingyi int mid = l + r >> 1, ls = p << 1, rs = p << 1 | 1
#define y0 y_csyakioi_0
#define y1 y_csyakioi_1
#define rep(i, x, y) for(int i = x; i <= y; ++i)
#define per(i, x, y) for(int i = x; i >= y; --i)
#define repg(i, u) for(int i = head[u]; i; i = e[i].nxt)
inline int read(){
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); }
	while(ch >= '0' && ch <= '9'){ x = x * 10 + (ch ^ 48); ch = getchar(); }
	return x * f;
}
const int N = 500010;
set <int> g[N], st[N];
set <int> ::iterator it;
int n, m, x, y, tot, cnt, ans1[N], ans2[N], idx[N], f[N];
inline void mp(int x, int y){
	f[idx[x] = idx[y] = ++cnt] = x;
	st[cnt].insert(x); st[cnt].insert(y);
}
inline void mian(){
	n = read(); m = read(); tot = 0; cnt = 0;
	rep(i, 1, n) g[i].clear(), idx[i] = 0;
	rep(i, 1, m) x = read(), y = read(), g[x].insert(y), g[y].insert(x);
	rep(i, 1, n){
		if(g[i].size() == n - 1){ ans1[i] = ans2[i] = ++tot; continue; }
		if(idx[i]) continue; x = 1; g[i].insert(n + 1);
		for(it = g[i].begin(); it != g[i].end(); ++it){
			int v = (*it); if(i == x) ++x;
			if(v != x) break; ++x;
		}
		if(!idx[x]){ mp(i, x); continue; } y = idx[x];
		if(st[y].size() == 2 || f[y] == x){
			st[y].insert(i);
			f[idx[i] = y] = x;
		}
		else st[y].erase(x), mp(i, x);
	}
	rep(i, 1, cnt){
		ans1[f[i]] = ++tot;
		for(it = st[i].begin(); it != st[i].end(); ++it){
			int j = (*it);
			if(j != f[i]) ans2[j] = tot, ans1[j] = ++tot;
		}
		ans2[f[i]] = tot;
		st[i].clear(); f[i] = 0;
	}
	rep(i, 1, n) printf("%d ", ans1[i]); puts("");
	rep(i, 1, n) printf("%d ", ans2[i]); puts("");
}
int main(){ int qwq = read(); while(qwq--) mian(); return 0; }