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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const int maxn = 2e5 + 5;
int T, n, m, k, tmp[maxn];
int num[maxn], vis[maxn];
vector<int> g[maxn], ans2;
unordered_set<int> mp[maxn];
set<pair<int, int> > st;
int gao() {
  while (st.size()) {
    int u = (*st.begin()).second;
    if (num[u] >= k) return 1;
    if (num[u] == k - 1) {
      int f = 1, cnt = 0;
      tmp[++cnt] = u;
      for (auto v : g[u])
        if (!vis[v]) tmp[++cnt] = v;
      for (register int i = 1; i <= cnt; ++i) {
        for (register int j = 1; j <= cnt; ++j)
          if (!mp[tmp[i]].count(tmp[j])) {
            f = 0;
            break;
          }
        if (!f) break;
      }
      if (f) {
        ans2.clear();
        for (int i = 1; i <= cnt; i++) ans2.push_back(tmp[i]);
        return 2;
      }
    }
    vis[u] = 1;
    st.erase({num[u], u});
    for (auto v : g[u]) {
      if (vis[v]) continue;
      st.erase({num[v], v});
      st.insert({--num[v], v});
    }
  }
  return 0;
}
signed main() {
  scanf("%d", &T);
  while (T--) {
    st.clear();
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i++) {
      mp[i].clear();
      mp[i].insert(i);
      num[i] = vis[i] = 0, g[i].clear();
    }
    for (int i = 1, u, v; i <= m; i++) {
      scanf("%d%d", &u, &v);
      g[u].push_back(v);
      g[v].push_back(u);
      num[u]++, num[v]++;
      mp[u].insert(v);
      mp[v].insert(u);
    }
    for (int i = 1; i <= n; i++) st.insert({num[i], i});
    int f = gao();
    if (f == 0) puts("-1");
    if (f == 1) {
      printf("%d %d\n", 1, st.size());
      for (auto i : st) printf("%d ", i.second);
      puts("");
    }
    if (f == 2) {
      puts("2");
      for (auto i : ans2) printf("%d ", i);
      puts("");
    }
  }
}