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 arriz=5.5e5;
struct info { int idx,val; };
bool operator <(info a,info b)
{
    return a.val==b.val?a.idx<b.idx:a.val<b.val;
}
struct stars { int rt; vector<int> leave; };
set<int> rv,g[arriz],t[arriz];
set<info> d;
int T,n,m,deg[arriz],perm1[arriz],perm2[arriz];
vector<stars>res;
void dfs(int u)
{
    rv.erase(u);
    for (int v,cnt=0; true; cnt=v)
    {
        auto iter=rv.upper_bound(cnt);
        if (iter==rv.end()) break;
        v=*iter;
        if (g[u].find(v)!=g[u].end()) continue;
        t[u].insert(v),t[v].insert(u),dfs(v);
    }
}
signed main()
{
    for (scanf("%d",&T); T--;)
    {
        scanf("%d %d",&n,&m);
        for (int i=1; i<=n; ++i)
            g[i].clear(),t[i].clear();
        rv.clear(),d.clear(),res.clear();
        for (int i=1,u,v; i<=m; ++i)
        {
            scanf("%d %d",&u,&v);
            g[u].insert(v),g[v].insert(u);
        }
        for (int i=1; i<=n; ++i) rv.insert(i);
        for (; !rv.empty();) dfs(*rv.begin());
        int rem=n;
        for (int i=1; i<=n; ++i)
        {
            deg[i]=t[i].size();
            if (deg[i]) d.insert({ i,deg[i] });
            else perm1[i]=perm2[i]=rem--;
        }
        for (int idx,f; !d.empty();)
        {
            idx=(*d.begin()).idx,f=*t[idx].begin();
            vector<int>leaves; leaves.clear();
            d.erase({ f,deg[f] });
            for (auto c:t[f])
            {
                d.erase({ c,deg[c] });
                if (deg[c]==1) leaves.push_back(c);
                else deg[c]--,d.insert({ c,deg[c] }),t[c].erase(f);
            }
            res.push_back({ f,leaves });
        }
        int l=0,r=0;
        for (auto c:res)
        {
            perm1[c.rt]=++l;
            for (auto ls:c.leave) perm1[ls]=++l,perm2[ls]=++r;
            perm2[c.rt]=++r;
        }
        for (int i=1; i<=n; ++i) printf("%d%c",perm1[i]," \n"[i==n]);
        for (int i=1; i<=n; ++i) printf("%d%c",perm2[i]," \n"[i==n]);
    }
    return 0;
}