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 pb push_back
#define mp make_pair
#define fs first
#define sc second
#define sz(a) ((int) (a).size())
#define all(x) (x).begin(), (x).end()

const long double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = acos(-1);

template <class T> T sqr (T x) {return x * x;}
template <class T> T abs (T x) {return x < 0 ? -x : x;}

template <class T>
vector<T> ReadVector(int n) {
    vector<T> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }
    return a;
}

template <class T>
void WriteVector(const vector<T>& a) {
    for (int i = 0; i < sz(a); ++i) {
        cout << a[i] << (i + 1 < sz(a) ? ' ' : '\n');
    }
}

void Solve() {
    int n, m;
    cin >> n >> m;

    vector<vector<int>> adj(n);
    for (int i = 0; i < m; ++i) {
        int v1, v2;
        cin >> v1 >> v2;
        --v1, --v2;
        adj[v1].pb(v2);
        adj[v2].pb(v1);
    }

    vector<int> good(n);
    for (int i = 0; i < n; ++i) {
        if (sz(adj[i]) == n - 1) {
            good[i] = -1;
            continue;
        }
        
        adj[i].pb(i);
        sort(all(adj[i]));
        
        good[i] = 0;
        while (good[i] < sz(adj[i]) && adj[i][good[i]] == good[i]) {
            ++good[i];
        }
    }

    vector<vector<int>> gr(n);
    vector<pair<int, int>> ind(n);
    for (int i = 0; i < n; ++i) {
        gr[i].pb(i);
        ind[i] = mp(i, 0);
    }

    for (int i = 0; i < n; ++i) {
        if (good[i] == -1) {
            continue;
        }

        if (sz(gr[ind[i].fs]) > 1) {
            continue;
        }

        int v = good[i];
        int g = ind[v].fs;
        if (sz(gr[g]) == 2 && ind[v].sc != 0) {
            swap(ind[gr[g][0]], ind[gr[g][1]]);
            swap(gr[g][0], gr[g][1]);
        }
        
        if (ind[v].sc == 0) {
            gr[ind[i].fs].pop_back();
            gr[g].pb(i);
            ind[i] = mp(g, sz(gr[g]) - 1);
            continue;
        }

        int pl = ind[v].sc;
        swap(ind[v], ind[gr[g].back()]);
        swap(gr[g][pl], gr[g].back());
        gr[g].pop_back();
        ind[v] = mp(ind[i].fs, 1);
        gr[ind[i].fs].pb(v);
    }

    int cur = 0;
    vector<int> p1(n);
    vector<int> p2(n);
    for (int i = 0; i < sz(gr); ++i) {
        for (int j = 0; j < sz(gr[i]); ++j) {
            p1[gr[i][j]] = cur + j;
            p2[gr[i][j]] = cur + (j + sz(gr[i]) - 1) % sz(gr[i]);
        }
        cur += sz(gr[i]);
    }

    for (int i = 0; i < n; ++i) {
        cout << p1[i] + 1 << (i + 1 < n ? ' ' : '\n');
    }
    
    for (int i = 0; i < n; ++i) {
        cout << p2[i] + 1 << (i + 1 < n ? ' ' : '\n');
    }
}

int main () {
    ios_base::sync_with_stdio(0);
    std::cin.tie(nullptr);
    
    // clock_t start_clock = clock();
    
    // freopen("_input.txt", "rt", stdin);
    // freopen("_output.txt", "wt", stdout);

    int tc = 1;
    cin >> tc;
    for (int ti = 0; ti < tc; ++ti) {
        Solve();
    }
    
    // double total_seconds = (double) (clock() - start_clock) / CLOCKS_PER_SEC;
    // cerr << "Working time: " << total_seconds << "s." << endl;

    return 0;
}
