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 = 1e5 + 100;
unordered_set<int> has[maxn];
set<pair<int, int> > st;
vector<int> chk;
int t, n, m, k, del[maxn], d[maxn], vis[maxn];
vector<int> E[maxn];
bool solved;
inline int read() {
  int num = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    num = (num << 3) + (num << 1) + (ch ^ 48);
    ch = getchar();
  }
  return w ? -num : num;
}
bool check() {
  if (chk.size() != k) return 0;
  for (int i = 0; i < k; ++i) {
    for (int j = i + 1; j < k; ++j) {
      int u = chk[i], v = chk[j];
      if (!has[u].count(v)) return 0;
    }
  }
  return 1;
}
int main() {
  t = read();
  while (t--) {
    st.clear();
    solved = 0;
    n = read();
    m = read();
    k = read();
    for (int i = 1; i <= n; ++i)
      has[i].clear(), E[i].clear(), del[i] = d[i] = 0;
    for (int i = 0; i < m; ++i) {
      int u, v;
      u = read();
      v = read();
      E[u].push_back(v);
      E[v].push_back(u);
      d[u]++;
      d[v]++;
      has[u].insert(v);
      has[v].insert(u);
    }
    if (k > 500) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; ++i) st.insert({d[i], i});
    while (!st.empty()) {
      auto pii = *(st.begin());
      int deg = pii.first, u = pii.second;
      if (deg >= k) {
        printf("1 %d\n", st.size());
        for (auto& v : st) printf("%d ", v.second);
        puts("");
        solved = 1;
        break;
      } else if (deg < k - 1) {
        st.erase(st.begin());
        del[u] = 1;
        for (auto& v : E[u]) {
          if (!del[v]) {
            st.erase({d[v], v});
            st.insert({--d[v], v});
          }
        }
      } else {
        st.erase(st.begin());
        del[u] = 1;
        chk.clear();
        chk.push_back(u);
        for (auto& v : E[u]) {
          if (!del[v]) {
            chk.push_back(v);
            st.erase({d[v], v});
            st.insert({--d[v], v});
          }
        }
        if (check()) {
          puts("2");
          for (auto& v : chk) printf("%d ", v);
          puts("");
          solved = 1;
          break;
        }
        for (auto& v : E[u]) {
          if (!del[v]) {
            has[u].erase(v);
          }
        }
      }
    }
    if (!solved) puts("-1");
  }
}