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;
using ll = long long;
using ii = pair<int, int>;
const int N = 100005;
int n, m, k;
vector<int> adj[N];
int cnt[N];
int mark[N];
int pass;
int aux[N];
bool ta[N];
bool clique(int x) {
  pass++;
  mark[x] = pass;
  int id = 0;
  for (int i : adj[x])
    if (!ta[i]) {
      mark[i] = pass;
      aux[id++] = i;
    }
  for (int i = 0; i < id; i++) {
    for (int j = i + 1; j < id; j++) {
      if (!binary_search(adj[aux[i]].begin(), adj[aux[i]].end(), aux[j]))
        return 0;
    }
  }
  return 1;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d %d", &n, &m, &k);
    for (int i = 1; i <= n; i++) {
      adj[i].clear();
      cnt[i] = 0;
      ta[i] = 0;
    }
    for (int i = 0; i < m; i++) {
      int a, b;
      scanf("%d %d", &a, &b);
      adj[a].push_back(b);
      adj[b].push_back(a);
      cnt[b]++;
      cnt[a]++;
    }
    for (int i = 1; i <= n; i++) {
      sort(adj[i].begin(), adj[i].end());
    }
    priority_queue<ii> pq;
    for (int i = 1; i <= n; i++) {
      pq.push({-cnt[i], i});
    }
    bool ok = 0;
    while (1) {
      if (pq.empty()) break;
      int c, x;
      tie(c, x) = pq.top();
      c = -c;
      if (c != cnt[x]) {
        pq.pop();
        continue;
      }
      if (c >= k) {
        vector<int> ans;
        while (!pq.empty()) {
          int c, x;
          tie(c, x) = pq.top();
          c = -c;
          pq.pop();
          if (c != cnt[x]) {
            continue;
          }
          ans.push_back(x);
        }
        printf("1 %d\n", (int)ans.size());
        for (int i : ans) {
          printf("%d ", i);
        }
        printf("\n");
        ok = 1;
        break;
      }
      pq.pop();
      if (c == k - 1) {
        if (clique(x)) {
          printf("2\n");
          printf("%d", x);
          for (int i : adj[x])
            if (!ta[i]) {
              printf(" %d", i);
            }
          printf("\n");
          ok = 1;
          break;
        }
      }
      cnt[x] = 0;
      ta[x] = 1;
      for (int i : adj[x])
        if (cnt[i]) {
          cnt[i]--;
          pq.push({-cnt[i], i});
        }
    }
    if (!ok) {
      printf("-1\n");
    }
  }
  return 0;
}