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>
bool chkmin(T &x, T y) {
  return x > y ? x = y, 1 : 0;
}
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, k;
int du[100005], q[100005], vis[100005];
vector<int> adj[100005];
bool check(vector<int> &vec) {
  assert(vec.size() == k);
  sort(vec.begin(), vec.end());
  for (int i = 0; i < vec.size(); i++) {
    for (int j = 0; j < vec.size(); j++) {
      if (i == j) continue;
      auto it = lower_bound(adj[vec[i]].begin(), adj[vec[i]].end(), vec[j]);
      if (it == adj[vec[i]].end() || *it != vec[j]) return false;
    }
  }
  printf("2\n");
  for (auto r : vec) printf("%d ", r);
  printf("\n");
  return true;
}
int main() {
  int T = readint();
  while (T--) {
    n = readint();
    m = readint();
    k = readint();
    for (int i = 1; i <= n; i++) vis[i] = du[i] = 0, adj[i].clear();
    int x, y;
    for (int i = 1; i <= m; i++) {
      x = readint();
      y = readint();
      adj[x].push_back(y), adj[y].push_back(x);
      du[x]++, du[y]++;
    }
    for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
    int front = 0, rear = 0;
    for (int i = 1; i <= n; i++)
      if (du[i] < k) q[rear++] = i, vis[i] = 1;
    bool fl = 0;
    while (front < rear) {
      int t = q[front++];
      vis[t] = 2;
      if (du[t] < k - 1) {
        for (auto v : adj[t])
          if ((--du[v]) < k && !vis[v]) q[rear++] = v, vis[v] = 1;
      } else if (du[t] == k - 1) {
        vector<int> vec(0);
        vec.push_back(t);
        for (auto v : adj[t])
          if (vis[v] <= 1 && du[v] >= k - 1) vec.push_back(v);
        if (vec.size() == k && check(vec)) {
          fl = 1;
          break;
        }
        for (auto v : adj[t])
          if ((--du[v]) < k && !vis[v]) q[rear++] = v, vis[v] = 1;
      }
    }
    vector<int> vec(0);
    for (int i = 1; i <= n; i++)
      if (!vis[i]) vec.push_back(i);
    if (!fl && vec.size()) {
      printf("1 %d\n", vec.size());
      for (auto r : vec) printf("%d ", r);
      printf("\n");
    } else if (!fl)
      printf("-1\n");
  }
  return 0;
}