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;
unordered_set<int> valid;
void drop(vector<unordered_set<int>> &g, int u) {
  for (int v : g[u]) {
    g[v].erase(u);
  }
  g[u].clear();
  valid.erase(u);
}
bool check(vector<unordered_set<int>> &g, int u) {
  unordered_set<int> c = g[u];
  for (int v : c) {
    for (int w : c) {
      if (w != v && g[v].find(w) == g[v].end()) {
        return false;
      }
    }
  }
  return true;
}
int solve() {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  vector<unordered_set<int>> g(n);
  valid.clear();
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    --a, --b;
    g[a].insert(b);
    g[b].insert(a);
  }
  vector<pair<int, int>> v;
  for (int i = 0; i < n; ++i) {
    valid.insert(i);
  }
  while (1) {
    if (valid.size() < k) {
      printf("-1\n");
      return 0;
    }
    v.clear();
    for (int i : valid) {
      v.push_back({g[i].size(), i});
    }
    sort(v.begin(), v.end());
    if (v[0].first >= k) {
      printf("1 %d\n", valid.size());
      for (int w : valid) {
        printf("%d ", w + 1);
      }
      printf("\n");
      return 0;
    }
    for (auto p : v) {
      int u = p.second;
      if (g[u].size() < k - 1) {
        drop(g, u);
      }
      if (g[u].size() == k - 1) {
        if (check(g, u)) {
          printf("2\n");
          printf("%d ", u + 1);
          for (int w : g[u]) {
            printf("%d ", w + 1);
          }
          printf("\n");
          return 0;
        } else {
          drop(g, u);
        }
      }
      if (g[u].size() >= k) {
        break;
      }
    }
  }
}
int main() {
  int t;
  scanf("%d", &t);
  for (int i = 0; i < t; ++i) {
    solve();
  }
  return 0;
}