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:
#include<bits/stdc++.h>
#define endl '\n'
using namespace std;
const int N=5e5+10;

unordered_set<int>edg[N],ans[N];
vector<int>edg2[N];
int p[N],q[N],father[N],n,m;
void solve()
{
    cin>>n>>m;
    for(int i=0;i<m;i++)
    {
        int a,b;
        cin>>a>>b;
        edg[a].insert(b);
        edg[b].insert(a);
    }
    for(int i=1;i<=n;i++)
    {
        if(edg[i].size()!=n-1)father[i]=i;
        else father[i]=0;
    }

    for(int i=1;i<=n;i++)if(father[i]==i&&edg2[i].empty())
    {
        for(int j=1;j<=n;j++)if(j!=i&&father[j]==j)
        {
            if(!edg[i].count(j))
            {
//                cout<<i<<' '<<j<<endl;
                edg2[i].push_back(j);
                edg2[j].push_back(i);
                break;
            }
        }
    }

    for(int i=1;i<=n;i++)if(ans[i].empty()&&father[i]==i)
    {
        for(auto j:edg2[i])if(father[j]==j)
        {
            ans[i].insert(j);
            father[j]=i;
        }

        if(ans[i].empty())
        {
            int u=edg2[i][0];
            ans[father[u]].erase(u);
            father[u]=u;
            for(int v:edg2[u])if(ans[v].empty()&&father[v]==v)
            {
                ans[u].insert(v);
                father[v]=u;
            }
        }
//        cout<<i<<' '<<ans[i].size()<<endl;
    }
//    for(int i=1;i<=n;i++)cout<<father[i]<<endl;

    int tot=0;
    for(int i=1;i<=n;i++)
    {
        if(!father[i])
        {
            p[i]=q[i]=++tot;
        }
        else if(ans[i].size())
        {
            p[i]=++tot;
            for(auto j:ans[i])
            {
//                cout<<i<<' '<<j<<endl;
                q[j]=tot;
                p[j]=++tot;
            }
            q[i]=tot;
        }
    }

    for(int i=1;i<=n;i++)cout<<p[i]<<' ';cout<<endl;
    for(int i=1;i<=n;i++)cout<<q[i]<<' ';cout<<endl;
    for(int i=1;i<=n;i++)
    {
        edg[i].clear();
        edg2[i].clear();
        ans[i].clear();
    }

}
int main()
{
    ios::sync_with_stdio(0),cin.tie(0);
//    freopen("1.txt","r",stdin);
    int _;cin>>_;
    while(_--)
        solve();
    return 0;
}