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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
int T, d[100010], n, m, k, vis[100010];
vector<int> G[100010];
bool clique(int S) {
  static vector<int> t;
  t.clear();
  t.push_back(S);
  for (auto v : G[S]) {
    if (vis[v] == 2) continue;
    if (d[v] < k - 1) return false;
    t.push_back(v);
  }
  for (auto u : t) {
    for (auto v : t) {
      if (u == v) break;
      if (!binary_search(G[u].begin(), G[u].end(), v)) return false;
    }
  }
  printf("2\n");
  for (auto u : t) {
    printf("%d ", u);
  }
  printf("\n");
  return true;
}
void Solve() {
  n = read(), m = read(), k = read();
  queue<int> q;
  for (int i = 1; i <= n; ++i) {
    G[i].clear();
    d[i] = vis[i] = 0;
  }
  for (int i = 1; i <= m; ++i) {
    int u = read(), v = read();
    G[u].push_back(v);
    G[v].push_back(u);
    ++d[u], ++d[v];
  }
  for (int i = 1; i <= n; ++i) {
    sort(G[i].begin(), G[i].end());
    if (d[i] < k) vis[i] = 1, q.push(i);
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    vis[u] = 2;
    if (d[u] == k - 1 && 1LL * k * (k - 1) <= 2 * m) {
      if (clique(u)) return;
    }
    for (auto v : G[u]) {
      if (--d[v] < k && !vis[v]) vis[v] = 1, q.push(v);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (!vis[i]) {
      vector<int> ans;
      for (int u = 1; u <= n; ++u) {
        if (!vis[u]) ans.push_back(u);
      }
      printf("1 %d\n", (int)ans.size());
      for (auto x : ans) {
        printf("%d ", x);
      }
      printf("\n");
      return;
    }
  }
  printf("-1\n");
}
int main() {
  T = read();
  while (T--) {
    Solve();
  }
  return 0;
}