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 <bits/stdc++.h>
using namespace std;
#define erorp(x) cout<<#x<<"={"<<(x.F)<<" , "<<x.S<<"}"<<endl;
#define print(v,r) f(i,0,r) cout<<v[i]<<" "; cout<<endl;
#define Get(x,y) scanf("%I64d%I64d",&x,&y);
#define is_bit(x,y) (x%(1<<(y+1))>=(1<<y))
#define eror(x) cout<<#x<<'='<<(x)<<endl;
#define f_(i,a,b) for(int i=a;i>=b;i--)
#define Gett(x,y) scanf("%d%d",&x,&y);
#define f(i,a,b) for(int i=a;i<b;i++)
#define get(x) scanf("%I64d",&x);
#define gett(x) scanf("%d",&x);
#define maxm(a,b) a=max(a,b);
#define minm(a,b) a=min(a,b);
#define Add(x,y) x=(x+y)%mod
#define lst(x) x[x.size()-1]
#define sz(x) int(x.size())
#define mp make_pair
#define ll long long
#define pb push_back
#define S second
#define F first

const int N=5e5+99;

int t,n,m,c,c1,c2,p[N],q[N];
vector<int> adj[N];
set<int> g[N];

void solve(){
    cin>>n>>m;
    c=n,c1=1,c2=1;
    f(i,0,m){
        int u,v;
        Gett(u,v);
        adj[u].pb(v);
        adj[v].pb(u);
    }
    f(i,1,n+1){
        int u=0;
        adj[i].pb(0);
        adj[i].pb(i);
        sort(adj[i].begin(),adj[i].end());
        if(adj[i].size()==n+1) p[i]=c,q[i]=c--;
        else{
            f(j,1,adj[i].size())
                if(adj[i][j]!=adj[i][j-1]+1)
                    u=adj[i][j]-1;
            if(adj[i][adj[i].size()-1]!=n) u=n;
            //cout<<i<<" "<<u<<endl;
            g[i].insert(u);
            g[u].insert(i);
        }
    }
    f(i,1,n+1){
        vector<int> v;
        while(g[i].size()){
            int u=*g[i].begin();
            g[i].erase(u);
            g[u].erase(i);
            if(g[u].size()==0 || (g[i].size()==0 && v.size()==0))
                v.pb(u);
        }
        f(j,0,v.size())
            g[i].insert(v[j]),g[v[j]].insert(i);
    }
    f(i,1,n+1){
        if(1<g[i].size() || (g[i].size() && i<(*g[i].begin()) && g[(*g[i].begin())].size()==1)){
            p[i]=c1++;
            for(auto u:g[i])
                p[u]=c1++,q[u]=c2++;
            q[i]=c2++;
        }
    }
    f(i,1,n+1) cout<<p[i]<<" "; cout<<endl;
    f(i,1,n+1) cout<<q[i]<<" "; cout<<endl;
    f(i,1,n+1)
        g[i].clear(),adj[i].clear();
}

int main(){
    cin>>t;
    while(t--)
        solve();
}