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, deg[100005], vis[100005];
vector<int> e[100005];
set<pair<int, int> > sets;
set<pair<int, int> >::iterator it;
unordered_map<int, bool> mp[100005];
int q[100005], top;
int t;
inline int read() {
  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 << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
bool check(int x) {
  top = 0;
  q[++top] = x;
  for (int i = 0; i < e[x].size(); i++) {
    int y = e[x][i];
    if (!vis[y]) continue;
    q[++top] = y;
  }
  for (int i = 1; i <= top; i++)
    for (int j = i + 1; j <= top; j++)
      if (!mp[q[i]][q[j]]) return 0;
  return 1;
}
int flag = 0;
void mian() {
  sets.clear();
  for (int i = 1; i <= n; i++) mp[i].clear(), deg[i] = vis[i] = 0, e[i].clear();
  n = read(), m = read(), k = read();
  for (int i = 1, u, v; i <= m; i++) {
    u = read(), v = read();
    deg[u]++, deg[v]++;
    e[u].push_back(v), e[v].push_back(u);
    mp[u][v] = mp[v][u] = 1;
  }
  for (int i = 1; i <= n; i++) sets.insert(make_pair(deg[i], i)), vis[i] = 1;
  while (1) {
    if (sets.empty()) break;
    pair<int, int> tmp = *sets.begin();
    if (tmp.first == k - 1 && check(tmp.second)) {
      cout << 2 << '\n';
      for (int i = 1; i <= top; i++) cout << q[i] << ' ';
      cout << '\n';
      return;
    }
    if (tmp.first < k) {
      sets.erase(tmp);
      vis[tmp.second] = 0;
      for (int i = 0; i < e[tmp.second].size(); i++) {
        int v = e[tmp.second][i];
        if (!vis[v]) continue;
        pair<int, int> it2 = make_pair(deg[v], v);
        sets.erase(it2);
        deg[v]--;
        sets.insert(make_pair(deg[v], v));
      }
    }
    if (tmp.first >= k) {
      top = 0;
      for (it = sets.begin(); it != sets.end(); it++) q[++top] = (*it).second;
      cout << 1 << ' ' << top << '\n';
      for (int i = 1; i <= top; i++) cout << q[i] << ' ';
      cout << '\n';
      return;
    }
  }
  cout << -1 << '\n';
}
signed main() {
  t = read();
  for (int i = 1; i <= t; i++) {
    flag = i;
    mian();
  }
}