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 INF = 1e9;
const int MOD = 1e9 + 7;
unordered_set<int> adj[100100];
priority_queue<pair<int, int> > pq;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, m, k;
    scanf("%d %d %d", &n, &m, &k);
    for (int i = 1; i <= n; i++) {
      adj[i].clear();
    }
    while (m--) {
      int x, y;
      scanf("%d %d", &x, &y);
      adj[x].insert(y);
      adj[y].insert(x);
    }
    while (pq.size()) pq.pop();
    for (int i = 1; i <= n; i++) {
      pq.push({-(int)adj[i].size(), i});
    }
    while (1) {
      while (pq.size()) {
        if (adj[pq.top().second].size() == -pq.top().first) break;
        pq.pop();
      }
      if (pq.size() == 0) {
        printf("-1\n");
        break;
      }
      if (-pq.top().first >= k) {
        vector<int> ans;
        for (int i = 1; i <= n; i++) {
          if (adj[i].size()) {
            ans.push_back(i);
          }
        }
        printf("1 %d\n", ans.size());
        for (int it : ans) {
          printf("%d ", it);
        }
        printf("\n");
        break;
      }
      if (-pq.top().first == k - 1) {
        int now = pq.top().second;
        int ok = 1;
        for (auto it = adj[now].begin(); ok && it != adj[now].end(); it++) {
          auto it2 = it;
          for (it2++; ok && it2 != adj[now].end(); it2++) {
            if (adj[*it].find(*it2) == adj[*it].end()) {
              ok = 0;
            }
          }
        }
        if (ok) {
          printf("2\n");
          for (auto it : adj[now]) {
            printf("%d ", it);
          }
          printf("%d\n", now);
          break;
        }
      }
      int now = pq.top().second;
      pq.pop();
      for (int nxt : adj[now]) {
        adj[nxt].erase(now);
        pq.push({-(int)adj[nxt].size(), nxt});
      }
      adj[now].clear();
    }
  }
  return 0;
}