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>
using namespace std;
template <class T>
using v2d = vector<vector<T>>;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
const int maxN = 1e5 + 10;
int n, m, k, deg[maxN];
vector<int> adj[maxN], cliques[maxN], candidates;
bool flag[maxN], good[maxN], f[maxN];
vector<pair<int, int>> vec[maxN];
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= (int)(n); ++i) {
    adj[i].clear();
    cliques[i].clear();
    flag[i] = good[i] = f[i] = 0;
    vec[i].clear();
  }
  candidates.clear();
  for (int i = 0; i < (int)(m); ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  if (k > 500) {
    cout << -1 << "\n";
    return;
  }
  priority_queue<pair<int, int>, vector<pair<int, int>>,
                 greater<pair<int, int>>>
      pq;
  for (int i = 1; i <= (int)(n); ++i) {
    deg[i] = (int)adj[i].size();
    pq.emplace(deg[i], i);
  }
  while (!pq.empty()) {
    int g, u;
    tie(g, u) = pq.top();
    pq.pop();
    if (flag[u]) {
      continue;
    }
    if (g >= k) {
      vector<int> ans;
      for (int i = 1; i <= (int)(n); ++i) {
        if (!flag[i]) {
          ans.emplace_back(i);
        }
      }
      cout << 1 << " " << (int)ans.size() << "\n";
      for (auto &x : ans) {
        cout << x << " ";
      }
      cout << "\n";
      return;
    }
    flag[u] = 1;
    if (g == k - 1) {
      candidates.emplace_back(u);
      vector<int> &d = cliques[u];
      for (auto &v : adj[u]) {
        if (!flag[v]) {
          d.emplace_back(v);
        }
      }
      for (auto &v : d) {
        vec[u].emplace_back(v, u);
        for (auto &k : d) {
          if (k < v) {
            vec[v].emplace_back(k, u);
          }
        }
      }
      d.emplace_back(u);
    }
    for (auto &v : adj[u]) {
      if (!flag[v]) {
        deg[v]--;
        pq.emplace(deg[v], v);
      }
    }
  }
  for (auto &u : candidates) {
    good[u] = 1;
  }
  for (int u = 1; u <= (int)(n); ++u) {
    for (auto &v : adj[u]) {
      f[v] = 1;
    }
    for (auto &p : vec[u]) {
      good[p.second] &= f[p.first];
    }
    for (auto &v : adj[u]) {
      f[v] = 0;
    }
  }
  for (auto &u : candidates) {
    if (good[u]) {
      cout << 2 << "\n";
      for (auto &v : cliques[u]) {
        cout << v << " ";
      }
      cout << "\n";
      return;
    }
  }
  cout << -1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}