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=500010;
int T,n,m,deg[maxn],per1[maxn],per2[maxn];
set<int> g[maxn],gg[maxn];
set<int> s;

struct node{
    int id,val;
    node(int x,int y){
        id=x;
        val=y;
    }
    friend bool operator < (node s1,node s2){
        if(s1.val!=s2.val) return s1.val<s2.val;
        return s1.id<s2.id;
    }
};

struct star{
    int center;
    vector<int> leaves;
}t;

set<node> d;

void dfs(int u){
    s.erase(u);
    int cur=0;
    while(1){
        auto iter=s.upper_bound(cur);
        if(iter==s.end()) break;
        int v=*iter;
        cur=v;
        if(g[u].find(v)!=g[u].end()) continue;
        gg[u].insert(v);
        gg[v].insert(u);
        dfs(v);
    }
}

void solve(int& l,int& r){
    per1[t.center]=++l;
    for(int u:t.leaves){
        per1[u]=++l;
        per2[u]=++r;
    }
    per2[t.center]=++r;
}

int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%d %d",&n,&m);
        for(int i=1;i<=n;i++){
            g[i].clear();
            gg[i].clear();
        }
        for(int i=1;i<=m;i++){
            int u,v;
            scanf("%d %d",&u,&v);
            g[u].insert(v);
            g[v].insert(u);
        }
        s.clear();
        d.clear();
        for(int i=1;i<=n;i++) s.insert(i);
        while(s.size()){
            int u=*s.begin();
            dfs(u);
        }
        int remain=n;
        for(int i=1;i<=n;i++){
            deg[i]=gg[i].size();
            if(deg[i]){
                d.insert({i,deg[i]});
            }
            else{
                per1[i]=per2[i]=remain--;
            }
        }
        int l=0,r=0;
        while(d.size()){
            t.leaves.clear();
            int id=(*d.begin()).id;
            int u=*gg[id].begin();
            d.erase({u,deg[u]});
            t.center=u;
            for(auto v:gg[u]){
                d.erase({v,deg[v]});
                if(deg[v]==1) t.leaves.push_back(v);
                else{
                    deg[v]--;
                    d.insert({v,deg[v]});
                    gg[v].erase(u);
                }
            }
            solve(l,r);
        }
        for(int i=1;i<n;i++) printf("%d ",per1[i]);
        printf("%d\n",per1[n]);
        for(int i=1;i<n;i++) printf("%d ",per2[i]);
        printf("%d\n",per2[n]);
    }
}

