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;
int n, m, k;
vector<int> adj[100005];
int din[100005];
queue<int> que;
int Vec[100005], V_size;
int a, b, c, d, cur;
bool succ;
int main() {
  int Test;
  scanf("%d", &Test);
  while (Test--) {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i += 1) {
      adj[i].clear();
      din[i] = 0;
    }
    while (!que.empty()) que.pop();
    for (int i = 1; i <= m; i += 1) {
      scanf("%d%d", &a, &b);
      adj[a].push_back(b), adj[b].push_back(a);
      din[a]++, din[b]++;
    }
    if (k == 1) {
      puts("2");
      puts("1");
      continue;
    }
    for (int i = 1; i <= n; i += 1) sort(adj[i].begin(), adj[i].end());
    for (int i = 1; i <= n; i += 1)
      if (din[i] < (k - 1)) que.push(i);
    while (!que.empty()) {
      cur = que.front();
      que.pop();
      for (auto& i : adj[cur]) {
        din[cur]--, din[i]--;
        if (din[i] == (k - 2)) que.push(i);
      }
    }
    for (int i = 1; i <= n; i += 1)
      if (din[i] == (k - 1)) que.push(i);
    bool nex_loop = false;
    while (!que.empty()) {
      cur = que.front();
      que.pop();
      if (din[cur] == (k - 1)) {
        V_size = 0;
        for (auto& i : adj[cur]) {
          if (din[i] <= 0) continue;
          Vec[V_size++] = i;
        }
        succ = true;
        for (int i = 0; i <= V_size - 1; i += 1) {
          for (int j = i + 1; j <= V_size - 1; j += 1) {
            a = Vec[i], b = Vec[j];
            auto it = lower_bound(adj[a].begin(), adj[a].end(), b);
            if ((it == adj[a].end()) || (*it != b)) {
              succ = false;
              break;
            }
          }
          if (!succ) break;
        }
        if (succ) {
          puts("2");
          printf("%d ", cur);
          for (int i = 0; i <= V_size - 1; i += 1) printf("%d ", Vec[i]);
          printf("\n");
          nex_loop = true;
          break;
        }
      }
      for (auto& i : adj[cur]) {
        if (din[i] <= 0) continue;
        din[cur]--, din[i]--;
        if (din[i] == (k - 1)) que.push(i);
      }
    }
    if (nex_loop) continue;
    V_size = 0;
    for (int i = 1; i <= n; i += 1) {
      if (din[i] <= 0) continue;
      Vec[V_size++] = i;
    }
    if (V_size == 0)
      puts("-1");
    else {
      printf("1 %d\n", V_size);
      for (int i = 0; i <= V_size - 1; i += 1) {
        printf("%d ", Vec[i]);
      }
      printf("\n");
    }
  }
  return 0;
}