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 pii = pair<int, int>;
const int NN = 101010;
const int MM = 4020;
int id[NN];
int edge[MM][MM];
int deg[NN];
vector<int> adj[NN];
int run;
int n, m, nn, k;
int pos[NN], flag[NN];
bool is_clique(int u, int k, vector<int> res) {
  run++;
  res.push_back(u);
  int tot = 0;
  for (int i = 0; i < k; i++) {
    int u = res[i];
    if (!pos[u])
      for (int v : adj[u]) {
        if (flag[v] == run) tot++;
      }
    flag[u] = run;
  }
  for (int i = 0; i < k; i++) {
    int u = res[i];
    if (pos[u]) {
      for (int j = i + 1; j < k; j++) {
        int v = res[j];
        if (pos[v]) {
          if (edge[pos[u]][pos[v]]) tot++;
        }
      }
    }
  }
  if (tot == 1LL * k * (k - 1) / 2) return 1;
  return 0;
}
void init() {
  for (int i = 1; i <= n; i++) {
    if (pos[i]) {
      for (int j : adj[i]) {
        if (pos[j]) {
          edge[pos[i]][pos[j]] = 0;
        }
      }
    }
  }
}
void Solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    adj[i].clear();
    id[i] = i;
    pos[i] = 0;
    deg[i] = 0;
  }
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    deg[u]++;
    adj[v].push_back(u);
    deg[v]++;
  }
  if (1LL * k * (k - 1) > m * 2) {
    puts("-1");
    init();
    return;
  }
  sort(id + 1, id + n + 1, [&](int u, int v) { return deg[u] > deg[v]; });
  nn = min(n, MM - 2);
  for (int i = 1; i <= nn; i++) pos[id[i]] = i;
  for (int i = 1; i <= n; i++) {
    if (pos[i]) {
      for (int j : adj[i]) {
        if (pos[j]) {
          edge[pos[i]][pos[j]] = 1;
        }
      }
    }
  }
  set<pii> st;
  for (int i = 1; i <= n; i++) {
    st.insert(pii(deg[i], i));
  }
  while (!st.empty()) {
    auto p = *st.begin();
    int u = p.second;
    if (p.first < k - 1) {
      st.erase(p);
      for (int v : adj[u]) {
        auto q = st.find(pii(deg[v], v));
        if (q != st.end()) {
          st.erase(q);
          deg[v]--;
          if (deg[v]) st.insert(pii(deg[v], v));
        }
      }
    } else if (p.first == k - 1) {
      vector<int> res;
      for (int v : adj[u]) {
        if (st.count(pii(deg[v], v))) res.push_back(v);
      }
      if (is_clique(u, k, res)) {
        puts("2");
        for (int v : adj[u]) {
          if (st.find(pii(deg[v], v)) != st.end()) printf("%d ", v);
        }
        printf("%d\n", u);
        init();
        return;
      } else {
        st.erase(p);
        for (int v : adj[u]) {
          auto q = st.find(pii(deg[v], v));
          if (q != st.end()) {
            st.erase(q);
            deg[v]--;
            if (deg[v]) st.insert(pii(deg[v], v));
          }
        }
      }
    } else {
      printf("1 %d\n", st.size());
      for (auto p : st) {
        printf("%d ", p.second);
      }
      puts("");
      init();
      return;
    }
  }
  puts("-1");
  init();
}
int main() {
  srand(time(0));
  int t;
  cin >> t;
  while (t--) Solve();
  return 0;
}