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 X first
#define Y second
const int N = 5e5;
set <int> st;
set <int> g[N];
set <int> t[N];
void dfs(int u)
{
    st.erase(u);
    int pos = -1;
    while(true)
    {
        auto it = st.upper_bound(pos);
        if(it == st.end())
        {
            break;
        }
        int v = *it;
        pos = v;
        if(g[u].find(pos) != g[u].end())
        {
            continue;
        }
        t[u].insert(v);
        t[v].insert(u);
        dfs(v);
    }
}
signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int q;
    cin >> q;
    while(q--)
    {
        int n, m;
        cin >> n >> m;
        st.clear();
        vector <int> a(n), b(n);
        for(int i = 0; i < n; i++)
        {
            g[i].clear();
            t[i].clear();
            st.insert(i);
        }
        vector <int> deg(n);
        for(int i = 0; i < m; i++)
        {
            int u, v;
            cin >> u >> v;
            u--;
            v--;
            g[u].insert(v);
            g[v].insert(u);
        }
        while(st.size() != 0)
        {
            int v = *st.begin();
            dfs(v);
        }
        set <pair <int, int> > st1;
        int it = n;
        for(int i = 0; i < n; i++)
        {
            deg[i] = t[i].size();
            if(deg[i] == 0)
            {
                a[i] = b[i] = it;
                it--;
            }
            else
            {
                st1.insert({deg[i], i});
            }
        }
        vector <pair <int, vector <int> > > res;
        while(st1.size() > 0)
        {
            int v = st1.begin() -> second;
            int u = *t[v].begin();
            st1.erase({t[u].size(), u});
            vector <int> leaves;
            for(auto to : t[u])
            {
                st1.erase({deg[to], to});
                if(deg[to] == 1)
                {
                    leaves.push_back(to);
                }
                else
                {
                    deg[to]--;
                    st1.insert({deg[to], to});
                    t[to].erase(u);
                }
            }
            res.push_back({u, leaves});
        }
        int l, r;
        l = r = 0;
        for(int i = 0; i < res.size(); i++)
        {
            a[res[i].X] = ++l;
            for(auto v : res[i].Y)
            {
                a[v] = ++l;
                b[v] = ++r;
            }
            b[res[i].X] = ++r;
        }
        for(int i = 0; i < n; i++)
        {
            cout << a[i] << " ";
        }
        cout << "\n";
        for(int i = 0; i < n; i++)
        {
            cout << b[i] << " ";
        }
        cout << "\n";
    }
    return 0;
}
