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>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
template <typename T, typename... Args>
void read(T &first, Args &...args) {
  read(first);
  read(args...);
}
template <typename T>
void write(T arg) {
  T x = arg;
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
template <typename T, typename... Ts>
void write(T arg, Ts... args) {
  write(arg);
  if (sizeof...(args) != 0) {
    putchar(' ');
    write(args...);
  }
}
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int N = 1e5 + 5;
int T;
int n, m, k, flag = 1;
vector<int> edge[N];
queue<int> q;
int vis[N], vis2[N], deg[N];
vector<int> ve;
void solve() {
  while (!q.empty()) q.pop();
  for (int i = 1; i <= n; i++)
    if (deg[i] < k) q.push(i), vis[i] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    vis[u] = 1;
    if (deg[u] == k - 1) {
      ve.clear();
      ve.push_back(u);
      for (auto it : edge[u]) {
        if (!vis2[it]) ve.push_back(it);
      }
      int mark = 1;
      for (int i = 0; i < ve.size(); i++) {
        for (int j = 0; j < ve.size(); j++) {
          if (ve[i] == ve[j]) break;
          if (!binary_search(edge[ve[i]].begin(), edge[ve[i]].end(), ve[j]))
            mark = 0;
        }
      }
      if (mark) {
        flag = 0;
        break;
      }
      ve.clear();
    }
    vis2[u] = 1;
    for (auto it : edge[u]) {
      deg[it]--;
      if (deg[it] < k && !vis[it]) {
        q.push(it), vis[it] = 1;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  while (T--) {
    flag = 1;
    cin >> n >> m >> k;
    for (int i = 1; i <= m; i++) {
      int u, v;
      cin >> u >> v;
      edge[u].push_back(v);
      edge[v].push_back(u);
    }
    for (int i = 1; i <= n; i++)
      sort(edge[i].begin(), edge[i].end()), deg[i] = (int)edge[i].size();
    solve();
    if (!flag) {
      cout << 2 << endl;
      for (auto it : ve) cout << it << ' ';
      cout << endl;
    } else {
      set<int> s;
      for (int i = 1; i <= n; i++) {
        for (auto it : edge[i])
          if (!vis[it]) s.insert(it);
      }
      if ((int)s.size() == 0)
        cout << -1 << endl;
      else {
        cout << 1 << ' ' << (int)s.size() << endl;
        for (auto it : s) cout << it << ' ';
        cout << endl;
      }
    }
    for (int i = 1; i <= n; i++) edge[i].clear(), deg[i] = vis[i] = vis2[i] = 0;
  }
}