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 NMAX = 1e5;
int t;
int n, m, k;
vector<int> graph[NMAX + 5];
int gr[NMAX + 5];
bool active[NMAX + 5];
bool in[NMAX + 5];
bool exists(int i, int j) {
  vector<int>::iterator it = lower_bound(graph[i].begin(), graph[i].end(), j);
  if (it != graph[i].end() && *it == j) {
    return true;
  }
  return false;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d %d", &n, &m, &k);
    for (int i = 1; i <= n; i++) {
      gr[i] = 0;
      graph[i].clear();
      active[i] = 1;
      in[i] = 0;
    }
    for (int i = 1; i <= m; i++) {
      int u, v;
      scanf("%d %d", &u, &v);
      graph[u].push_back(v);
      graph[v].push_back(u);
      gr[u]++;
      gr[v]++;
    }
    if (1LL * k * (k - 1) > 2 * m) {
      printf("-1\n");
      continue;
    }
    queue<int> q;
    for (int i = 1; i <= n; i++) {
      sort(graph[i].begin(), graph[i].end());
      if (gr[i] < k) {
        q.push(i);
        in[i] = 1;
      }
    }
    while (!q.empty()) {
      int nod = q.front();
      active[nod] = 0;
      q.pop();
      for (auto it : graph[nod]) {
        gr[it]--;
        if (in[it] == 0 && gr[it] < k) {
          in[it] = 1;
          q.push(it);
        }
      }
    }
    vector<int> ans;
    for (int i = 1; i <= n; i++) {
      if (active[i] == 1) {
        ans.push_back(i);
      }
    }
    if (ans.empty() == false) {
      printf("1 %d\n", (int)ans.size());
      for (auto it : ans) {
        printf("%d ", it);
      }
      printf("\n");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      gr[i] = graph[i].size();
      active[i] = 1;
      in[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      if (gr[i] < k) {
        in[i] = 1;
        q.push(i);
      }
    }
    bool found = false;
    while (!q.empty()) {
      int nod = q.front();
      q.pop();
      active[nod] = 0;
      if (gr[nod] == k - 1) {
        vector<int> nodes = {nod};
        for (auto it : graph[nod]) {
          if (active[it] == 1) {
            nodes.push_back(it);
          }
        }
        bool ok = true;
        for (int i = 0; i < (int)nodes.size() && ok; i++) {
          for (int j = i + 1; j < (int)nodes.size() && ok; j++) {
            ok &= exists(nodes[i], nodes[j]);
          }
        }
        if (ok == true) {
          found = true;
          printf("2\n");
          for (auto it : nodes) {
            printf("%d ", it);
          }
          printf("\n");
          break;
        }
      }
      for (auto it : graph[nod]) {
        gr[it]--;
        if (in[it] == 0 && gr[it] < k) {
          in[it] = 1;
          q.push(it);
        }
      }
    }
    if (found == false) {
      printf("-1\n");
      continue;
    }
  }
  return 0;
}