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;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, m, k;
    scanf("%d %d %d", &n, &m, &k);
    vector<unordered_set<int, custom_hash> > g(n);
    while (m--) {
      int u, v;
      scanf("%d %d", &u, &v);
      u--, v--;
      g[u].insert(v);
      g[v].insert(u);
    }
    priority_queue<pair<int, int> > q;
    vector<bool> vis(n, false);
    for (int i = 0; i < n; i++) q.push(make_pair(-((int)g[i].size()), i));
    int res = -1;
    while (!q.empty()) {
      auto [deg, u] = q.top();
      q.pop();
      if (vis[u]) continue;
      vis[u] = true;
      deg = -deg;
      if (deg >= k) {
        vis[u] = false;
        res = 1;
        break;
      }
      if (deg == k - 1) {
        bool ok = true;
        vector<int> w;
        for (int v : g[u]) {
          for (int z : w) {
            if (!g[v].count(z)) {
              ok = false;
              break;
            }
          }
          if (!ok) break;
          w.push_back(v);
        }
        if (ok) {
          w.push_back(u);
          res = 2;
          printf("2\n");
          for (int i = 0; i < (int)w.size(); i++) {
            if (i > 0) printf(" ");
            printf("%d", w[i] + 1);
          }
          printf("\n");
          break;
        }
      }
      for (int v : g[u]) {
        g[v].erase(u);
        q.push(make_pair(-((int)g[v].size()), v));
      }
    }
    if (res == -1) {
      printf("-1\n");
      continue;
    } else if (res == 1) {
      vector<int> w;
      for (int i = 0; i < n; i++)
        if (!vis[i]) w.push_back(i);
      printf("1 %d\n", (int)w.size());
      for (int i = 0; i < (int)w.size(); i++) {
        if (i > 0) printf(" ");
        printf("%d", w[i] + 1);
      }
      printf("\n");
    }
  }
  return 0;
}