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 long long mod = 1e9 + 7;
const long long M = 1e8;
const long long N = 4e6 + 5;
vector<long long> e[N], c;
long long in[N], v[N], gap[N];
long long n, m, k;
void init() {
  for (long long i = 0; i <= n; i++) e[i].clear();
  c.clear();
  for (long long i = 1; i <= n; i++) in[i] = v[i] = 0;
}
void bfs() {
  long long tot = n;
  queue<long long> q;
  for (long long i = 1; i <= n; i++)
    if (in[i] < k) q.push(i);
  while (q.size()) {
    long long x = q.front();
    q.pop();
    if (v[x]) continue;
    tot--;
    v[x] = 1;
    if (in[x] == k - 1) {
      c.clear();
      c.push_back(x);
      long long sz = e[x].size();
      for (long long i = 0; i < sz; i++) {
        long long y = e[x][i];
        if (v[y]) continue;
        c.push_back(y);
      }
      long long flag = 0;
      sz = c.size();
      for (long long i = 0; i < sz; i++)
        for (long long j = 0; j < sz; j++) {
          long long y1 = c[i];
          long long y2 = c[j];
          if (y1 == y2) continue;
          if (!binary_search(e[y1].begin(), e[y1].end(), y2)) {
            flag = 1;
            break;
          }
        }
      if (!flag) {
        puts("2");
        long long sz = c.size();
        for (long long i = 0; i < sz; i++) printf("%lld ", c[i]);
        puts("");
        return;
      }
    }
    long long sz = e[x].size();
    for (long long i = 0; i < sz; i++) {
      long long y = e[x][i];
      in[y]--;
      if (v[y]) continue;
      if (in[y] < k) q.push(y);
    }
  }
  if (tot > 0) {
    printf("1 %lld\n", tot);
    for (long long i = 1; i <= n; i++)
      if (!v[i]) printf("%lld ", i);
    puts("");
  } else
    puts("-1");
}
void solve() {
  cin >> n >> m >> k;
  init();
  for (long long i = 1; i <= m; i++) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    e[x].push_back(y);
    e[y].push_back(x);
    in[x]++;
    in[y]++;
  }
  if (k * (k - 1) > m * 2) {
    puts("-1");
    return;
  }
  for (long long i = 1; i <= n; i++) sort(e[i].begin(), e[i].end());
  bfs();
}
signed main() {
  long long T = 1;
  cin >> T;
  for (long long index = 1; index <= T; index++) {
    solve();
  }
  return 0;
}