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;
template <typename T>
void read(T &t) {
  t = 0;
  char ch = getchar();
  int f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  do {
    (t *= 10) += ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  t *= f;
}
const int maxn = (1e5) + 10;
int T, n, m, k;
bool flag;
vector<int> ans;
unordered_set<int> g[maxn];
priority_queue<pair<int, int> > q;
unordered_set<int>::iterator it, it2;
int main() {
  int x, y;
  read(T);
  while (T--) {
    read(n), read(m), read(k);
    ans.clear();
    for (int i = 1; i <= n; i++) g[i].clear();
    for (int i = 1; i <= m; i++) {
      read(x), read(y);
      g[x].insert(y), g[y].insert(x);
    }
    if (k == 1) {
      printf("2\n1\n");
      continue;
    }
    q = priority_queue<pair<int, int> >();
    for (int i = 1; i <= n; i++) q.push(make_pair(-(int)g[i].size(), i));
    flag = 0;
    while (!q.empty()) {
      int u = q.top().second;
      q.pop();
      if (g[u].empty()) continue;
      if ((int)g[u].size() >= k) {
        flag = 1;
        break;
      }
      if ((int)g[u].size() == k - 1 && (long long)k * (k - 1) / 2 <= m) {
        bool f = 1;
        for (it = g[u].begin(); f && it != g[u].end(); it++)
          for (it2 = it, it2++; f && it2 != g[u].end(); it2++) {
            f &= g[*it].find(*it2) != g[*it].end();
          }
        if (f) {
          for (it = g[u].begin(); it != g[u].end(); it++) ans.push_back(*it);
          ans.push_back(u);
          break;
        }
      }
      for (it = g[u].begin(); it != g[u].end(); it++)
        g[*it].erase(u), q.push(make_pair(-(int)g[*it].size(), *it));
      g[u].clear();
    }
    if (ans.size()) {
      printf("2\n");
      for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
      printf("\n");
    } else if (flag) {
      for (int i = 1; i <= n; i++)
        if ((int)g[i].size()) ans.push_back(i);
      printf("1 %d\n", (int)ans.size());
      for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
      printf("\n");
    } else
      printf("-1\n");
  }
  return 0;
}