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 N = 1e5 + 20;
const int INF = 1e9 + 7;
int n, m, k;
queue<int> q;
unordered_set<int> g[N], ans;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void init() {
  for (register int i = 1, _r = n; i <= _r; ++i) g[i].clear();
  ans.clear();
  while (!q.empty()) q.pop();
}
int main() {
  int T = read();
  while (T--) {
    n = read(), m = read(), k = read();
    for (register int i = 1, _r = m; i <= _r; ++i) {
      int u = read(), v = read();
      g[u].insert(v);
      g[v].insert(u);
    }
    if (1LL * k * (k - 1) / 2 > m) {
      puts("-1");
      init();
      continue;
    }
    for (register int i = 1, _r = n; i <= _r; ++i)
      g[i].size() < k ? q.push(i) : (void)ans.insert(i);
    while (!q.empty()) {
      int st = q.front();
      q.pop();
      if (g[st].size() == k - 1) {
        int check = 1;
        for (int i : g[st])
          if (check)
            for (int j : g[st])
              if (g[i].find(j) == g[i].end() && i != j) {
                check = 0;
                break;
              }
        if (check) {
          puts("2");
          printf("%d", st);
          for (int i : g[st]) printf(" %d", i);
          puts("");
          ans.insert(0);
          break;
        }
      }
      for (int p : g[st]) {
        g[p].erase(st);
        if (g[p].size() + 1 == k) ans.erase(p), q.push(p);
      }
    }
    if (ans.find(0) == ans.end()) {
      if (ans.size()) {
        printf("1 %d\n", ans.size());
        for (int i : ans) printf("%d ", i);
        puts("");
      } else
        puts("-1");
    }
    init();
  }
  return 0;
}