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;
using ll = long long;
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<unordered_set<int>> g(n + 1);
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].insert(v);
    g[v].insert(u);
  }
  if (1LL * k * (k - 1) > 2 * m) {
    cout << "-1\n";
    return;
  }
  vector<int> deg(n + 1);
  for (int i = 1; i <= n; ++i) deg[i] = g[i].size();
  vector<char> used(n + 1);
  queue<int> que;
  int used_cnt = 0;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] < k) {
      que.emplace(i);
      used[i] = true;
      ++used_cnt;
    }
  }
  while (!que.empty()) {
    int v = que.front();
    que.pop();
    for (auto to : g[v]) {
      if (!used[to] && --deg[to] < k) {
        que.push(to);
        used[to] = true;
        ++used_cnt;
      }
    }
  }
  if (used_cnt != n) {
    cout << 1 << " " << n - used_cnt << "\n";
    for (int i = 1; i <= n; ++i) {
      if (!used[i]) cout << i << " ";
    }
    cout << "\n";
    return;
  }
  used_cnt = 0;
  fill(used.begin(), used.end(), 0);
  queue<int> clique_candidates;
  for (int i = 1; i <= n; ++i) {
    deg[i] = g[i].size();
    if (deg[i] < k - 1) {
      que.push(i);
      used[i] = true;
      ++used_cnt;
    }
    if (deg[i] == k - 1) {
      clique_candidates.emplace(i);
    }
  }
  while (used_cnt < n) {
    while (!que.empty()) {
      int v = que.front();
      que.pop();
      for (auto to : g[v]) {
        if (used[to]) continue;
        if (g[to].count(v)) {
          g[to].erase(v);
        }
        --deg[to];
        if (deg[to] == k - 1) {
          clique_candidates.emplace(to);
        }
        if (deg[to] < k - 1) {
          que.push(to);
          used[to] = true;
          ++used_cnt;
        }
      }
    }
    while (!clique_candidates.empty()) {
      auto v = clique_candidates.front();
      clique_candidates.pop();
      if (used[v]) continue;
      bool good = !g[v].empty();
      for (auto to : g[v]) {
        int have = 1;
        for (auto to2 : g[v]) {
          have += g[to].count(to2);
        }
        if (have != k - 1) {
          good = false;
          break;
        }
      }
      if (good) {
        cout << "2\n";
        cout << v;
        for (auto to : g[v]) {
          cout << " " << to;
        }
        cout << "\n";
        return;
      }
      que.emplace(v);
      used[v] = true;
      ++used_cnt;
    }
  }
  cout << "-1\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}