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("Ofast", "unroll-loops", "no-stack-protector")
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using uint = unsigned int;
using ull = unsigned long long;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
default_random_engine generator;
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(12) << fixed;
  cerr << setprecision(12) << fixed;
  int tests = 1;
  cin >> tests;
  for (int test = 1; test <= tests; ++test) {
    solve();
  }
}
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<int>> g(n);
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  set<pii> setik;
  vector<int> order;
  vector<int> t(n);
  vector<int> used(n);
  vector<int> deg(n);
  for (int i = 0; i < n; ++i) {
    deg[i] = (int)(g[i]).size();
    setik.insert({deg[i], i});
  }
  for (int i = 0; !setik.empty(); ++i) {
    auto it = setik.begin();
    auto [d, v] = *it;
    if (d >= k) {
      break;
    }
    setik.erase(it);
    used[v] = true;
    order.push_back(v);
    t[v] = i;
    for (int u : g[v]) {
      if (!used[u]) {
        setik.erase({deg[u], u});
        --deg[u];
        setik.insert({deg[u], u});
      }
    }
  }
  if (!setik.empty()) {
    vector<int> res;
    for (int i = 0; i < n; ++i) {
      if (!used[i]) {
        res.push_back(i);
      }
    }
    cout << 1 << " " << (int)(res).size() << "\n";
    for (int v : res) {
      cout << v + 1 << " ";
    }
    cout << "\n";
    return;
  }
  for (int v : order) {
    vector<int> gg;
    for (int u : g[v]) {
      if (t[v] < t[u]) {
        gg.push_back(u);
      }
    }
    g[v] = gg;
  }
  vector<int> cnt(n);
  for (int v : order) {
    if ((int)(g[v]).size() != k - 1) {
      continue;
    }
    for (int u : g[v]) {
      for (int w : g[u]) {
        ++cnt[w];
      }
    }
    ll sum = 0;
    for (int u : g[v]) {
      sum += cnt[u];
    }
    for (int u : g[v]) {
      for (int w : g[u]) {
        cnt[w] = 0;
      }
    }
    if (sum == ll(k - 2) * (k - 1) / 2) {
      cout << 2 << "\n";
      cout << v + 1 << " ";
      for (int u : g[v]) {
        cout << u + 1 << " ";
      }
      cout << "\n";
      return;
    }
  }
  cout << -1 << "\n";
}