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;
int32_t main()
{
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        cin>>n>>m;
        vector <int> a[n];
        int deq[n];
        for(int i=0;i<n;++i) deq[i]=0;
        set <pair<int,int> > ed;
        for(int i=0;i<m;++i)
        {
            int x,y;
            cin>>x>>y;
            x--;y--;
            a[x].push_back(y);a[y].push_back(x);
            deq[x]++;deq[y]++;
            ed.insert({x,y});ed.insert({y,x});
        }
        int p[n],q[n];
        set <pair<int,int> > o;
        for(int i=0;i<n;++i) o.insert({-deq[i],i});
        int sz=n;
        int curr=1;
        set <int> ok;
        for(int i=0;i<n;++i) ok.insert(i);
        while(sz!=0)
        {
            pair <int,int> h=(*o.begin());
            if((-h.first)==(sz-1))
            {
                p[h.second]=curr;q[h.second]=curr;
                ++curr;
                sz--;
                o.erase(o.begin());
                ok.erase(h.second);
                for(auto h:ok)
                {
                    o.erase({-deq[h],h});
                    deq[h]--;
                    o.insert({-deq[h],h});
                }
            }
            else
            {
                int v=h.second;
                //cout<<v<<" v "<<endl;
                int v2=(-1);
                for(auto h:ok)
                {
                    if(h!=v && !ed.count({v,h}))
                    {
                        v2=h;
                        break;
                    }
                }
                //cout<<v2<<" v2 "<<endl;
                vector <int> z;
                z.push_back(v);
                if(deq[v]==(sz-2))
                {
                for(auto h:ok)
                {
                    if(h!=v && h!=v2 && !ed.count({v2,h}) && deq[h]==(sz-2))
                    {
                        z.push_back(h);
                    }
                }
                }
                else if(deq[v]==(sz-3))
                {
                for(auto h:ok)
                {
                    if(h!=v && h!=v2 && !ed.count({v2,h}) && !ed.count({v,h}) && deq[h]==(sz-3))
                    {
                        z.push_back(h);
                    }
                }
                }
                for(int i=0;i<z.size();++i)
                {
                    p[z[i]]=(curr+i);q[z[i]]=(curr+i+1);
                }
                p[v2]=(curr+z.size());q[v2]=curr;
                z.push_back(v2);
                //cout<<" z "<<endl;
                //for(auto h:z) cout<<h<<' ';
                //cout<<endl;
                curr+=z.size();
                for(auto h:z) {o.erase({-deq[h],h});ok.erase(h);}
                sz=ok.size();
                for(auto h:z)
                {
                    for(auto i:a[h])
                    {
                        if(ok.count(i))
                        {
                            o.erase({-deq[i],i});
                            deq[i]--;
                            o.insert({-deq[i],i});
                        }
                    }
                }
            }
        }
        for(int i=0;i<n;++i) cout<<p[i]<<' ';
        cout<<endl;
        for(int i=0;i<n;++i) cout<<q[i]<<' ';
        cout<<endl;
    }
    return 0;
}
