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;
 
const int maxn=500007;
 
struct info{
    int idx,val;
    friend bool operator<(info u,info v){
        return u.val==v.val?u.idx<v.idx:u.val<v.val;
    }
};
 
struct stars{
    int rt;
    vector<int> leaves;
};
set<int> rv; // remained vertices
set<int> g[maxn]; // original graph
set<int> t[maxn]; // dfs tree
set<info> d;
int deg[maxn];
int n,m;
int perm1[maxn],perm2[maxn];
vector<stars> res;
 
void dfs(int u){
    rv.erase(u);
    int crt=0;
    while (1){
        auto iter=rv.upper_bound(crt);
        if (iter==rv.end()) break;
        int v=*iter;
        crt=v;
        if (g[u].find(v)!=g[u].end()) continue;
        t[u].insert(v), t[v].insert(u);
        dfs(v);
    }
}
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int tc;
    cin>>tc;
    while (tc--){
        cin>>n>>m;
        for (int i=1;i<=n;++i) g[i].clear(), t[i].clear();
        rv.clear(), res.clear(), d.clear();
        for (int i=1;i<=m;++i){
            int u,v;
            cin>>u>>v;
            g[u].insert(v), g[v].insert(u);
        }
        for (int i=1;i<=n;++i) rv.insert(i);
        while (rv.size()>0){
            int u=*(rv.begin());
            dfs(u);       
        }
        int rem=n;
        for (int i=1;i<=n;++i){
            deg[i]=t[i].size();
            if (deg[i]){
                d.insert((info){i,deg[i]});
            }
            else{
                perm1[i]=perm2[i]=rem;
                rem--;
            }
        }
        while (d.size()){
            int idx=(*(d.begin())).idx;
            int f=*(t[idx].begin());
            vector<int> leaves;
            leaves.clear();
            d.erase((info){f,deg[f]});
            for (auto c:t[f]){
                d.erase((info){c,deg[c]});
                if (deg[c]==1) leaves.push_back(c);
                else deg[c]--, d.insert((info){c,deg[c]}), t[c].erase(f);
            }
            res.push_back((stars){f,leaves});
        }
        int l=0,r=0;
        for (auto c:res){
            perm1[c.rt]=++l;
            for (auto ls:c.leaves){
                perm1[ls]=++l;
                perm2[ls]=++r;
            }
           perm2[c.rt]=++r;
        }
        for (int i=1;i<=n;++i) cout<<perm1[i]<<" ";
        cout<<"\n";
        for (int i=1;i<=n;++i) cout<<perm2[i]<<" ";
        cout<<"\n";
    }
}
