Problem: You are given an undirected graph with n vertices and m edges. Also, you are given an integer k.

Find either a clique of size k or a non-empty subset of vertices such that each vertex of this subset has at least k neighbors in the subset. If there are no such cliques and subsets report about it.

A subset of vertices is called a clique of size k if its size is k and there exists an edge between every two vertices from the subset. A vertex is called a neighbor of the other vertex if there exists an edge between them.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The next lines contain descriptions of test cases.

The first line of the description of each test case contains three integers n, m, k (1 ≤ n, m, k ≤ 10^5, k ≤ n).

Each of the next m lines contains two integers u, v (1 ≤ u, v ≤ n, u ≠ v), denoting an edge between vertices u and v.

It is guaranteed that there are no self-loops or multiple edges. It is guaranteed that the sum of n for all test cases and the sum of m for all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case: 

If you found a subset of vertices such that each vertex of this subset has at least k neighbors in the subset in the first line output 1 and the size of the subset. On the second line output the vertices of the subset in any order.

If you found a clique of size k then in the first line output 2 and in the second line output the vertices of the clique in any order.

If there are no required subsets and cliques print -1.

If there exists multiple possible answers you can print any of them.

Example

Input


3
5 9 4
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
10 15 3
1 2
2 3
3 4
4 5
5 1
1 7
2 8
3 9
4 10
5 6
7 10
10 8
8 6
6 9
9 7
4 5 4
1 2
2 3
3 4
4 1
1 3


Output


2
4 1 2 3 
1 10
1 2 3 4 5 6 7 8 9 10 
-1

Note

In the first test case: the subset \{1, 2, 3, 4\} is a clique of size 4.

In the second test case: degree of each vertex in the original graph is at least 3. So the set of all vertices is a correct answer.

In the third test case: there are no cliques of size 4 or required subsets, so the answer is -1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
using namespace std;
int n, m, r;
vector<vector<int> > g(500000);
vector<set<int> > mn(500000);
set<pair<int, int> > s;
int siz[500000];
int p[500000];
bool check(vector<int> v) {
  for (int i = 0; i < v.size(); i++)
    for (int j = i + 1; j < v.size(); j++) {
      int l = 0, r = g[v[i]].size() - 1;
      while (l < r) {
        int mid = (l + r) / 2;
        if (l == r - 1) mid++;
        if (g[v[i]][mid] > v[j])
          r = mid - 1;
        else
          l = mid;
      }
      if (g[v[i]][l] != v[j]) return false;
    }
  return true;
}
vector<int> v;
bool solve1() {
  s.clear();
  long long now = 0;
  for (int i = 0; i < n; i++) {
    now += g[i].size();
    s.insert(make_pair(g[i].size(), i));
    siz[i] = g[i].size();
  }
  bool us[n];
  for (int i = 0; i < n; i++) us[i] = false;
  while (s.size() > 0) {
    if (s.size() < r) break;
    pair<int, int> k = *(s.begin());
    if ((r * (r - 1)) <= now && k.first == r - 1) {
      v.clear();
      v.push_back(k.second);
      p[k.second] = 1;
      for (int i = 0; i < g[k.second].size(); i++)
        if (!us[g[k.second][i]]) {
          p[g[k.second][i]] = 1;
          v.push_back(g[k.second][i]);
        }
      if (check(v)) {
        cout << 2 << "\n";
        for (int i = 0; i < v.size(); i++) cout << v[i] + 1 << ' ';
        cout << "\n";
        p[k.second] = 0;
        for (int i = 0; i < g[k.second].size(); i++)
          if (!us[g[k.second][i]]) p[g[k.second][i]] = 0;
        return true;
      }
      p[k.second] = 0;
      for (int i = 0; i < g[k.second].size(); i++)
        if (!us[g[k.second][i]]) p[g[k.second][i]] = 0;
    }
    if (k.first >= r) {
      vector<int> ans;
      for (int i = 0; i < n; i++)
        if (!us[i]) ans.push_back(i + 1);
      cout << "1 " << ans.size() << "\n";
      for (int i = 0; i < ans.size(); i++) cout << ans[i] << ' ';
      cout << "\n";
      return true;
    }
    us[k.second] = true;
    s.erase(s.begin());
    for (int i = 0; i < g[k.second].size(); i++) {
      if (!us[g[k.second][i]]) {
        s.erase(make_pair(siz[g[k.second][i]], g[k.second][i]));
        s.insert(make_pair(siz[g[k.second][i]] - 1, g[k.second][i]));
        siz[g[k.second][i]]--;
        now -= 2;
      }
    }
  }
  return false;
}
void solve() {
  cin >> n >> m >> r;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
    mn[x].insert(y);
    mn[y].insert(x);
  }
  for (int i = 0; i < n; i++) sort(g[i].begin(), g[i].end());
  bool k1 = solve1();
  if (k1) {
    for (int i = 0; i < n; i++) g[i].clear();
    return;
  }
  for (int i = 0; i < n; i++) g[i].clear();
  cout << -1 << "\n";
  return;
}
int main() {
  ios::sync_with_stdio(0);
  int q;
  cin >> q;
  while (q--) solve();
}