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;
const int maxn = 100100;
int N, M, K, deg[maxn];
bool used[maxn];
unordered_set<int> G[maxn];
bool check(int u) {
  for (int v : G[u]) {
    for (int w : G[u]) {
      if (v == w) continue;
      if (!G[v].count(w)) return 0;
    }
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> N >> M >> K;
    for (int i = 1; i <= N; ++i) {
      G[i].clear();
      deg[i] = used[i] = 0;
    }
    for (int u, v, i = 1; i <= M; ++i) {
      cin >> u >> v;
      G[u].insert(v);
      G[v].insert(u);
      deg[u]++;
      deg[v]++;
    }
    if (1ll * K * (K - 1) > 2ll * M) {
      cout << -1 << '\n';
      continue;
    }
    if (K == 1) {
      cout << 2 << '\n' << 1 << '\n';
      continue;
    }
    queue<int> Q;
    for (int i = 1; i <= N; ++i)
      if (deg[i] < K) {
        Q.push(i);
        used[i] = 1;
      }
    bool con = 0;
    while (!Q.empty()) {
      int u = Q.front();
      Q.pop();
      if (deg[u] == K - 1) {
        if (check(u)) {
          cout << 2 << '\n';
          for (int v : G[u]) cout << v << ' ';
          cout << u << '\n';
          con = 1;
          break;
        }
      }
      while (!G[u].empty()) {
        int v = *G[u].begin();
        G[v].erase(u);
        G[u].erase(v);
        deg[u]--;
        deg[v]--;
        if (deg[v] < K && !used[v]) {
          Q.push(v);
          used[v] = 1;
        }
      }
    }
    if (con) continue;
    vector<int> Ans;
    for (int i = 1; i <= N; ++i)
      if (!used[i]) Ans.push_back(i);
    if (Ans.empty()) {
      cout << -1 << '\n';
    } else {
      cout << 1 << ' ' << (int)Ans.size() << '\n';
      for (int u : Ans) cout << u << ' ';
      cout << '\n';
    }
  }
  return 0;
}