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;
using ll = long long;
const int maxn = 100010;
int t, n, m, k;
vector<int> adj[maxn];
int deg[maxn];
int odeg[maxn];
bool marked[maxn];
bool isin[maxn];
set<int> ed[maxn];
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    adj[i].clear();
    deg[i] = 0;
    odeg[i] = 0;
    marked[i] = isin[i] = false;
    ed[i].clear();
  }
  int u, v;
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
    deg[u]++;
    deg[v]++;
    ed[u].insert(v);
    ed[v].insert(u);
  }
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    odeg[i] = deg[i];
    if (deg[i] < k) q.push(i);
  }
  while (q.size()) {
    u = q.front();
    q.pop();
    marked[u] = true;
    for (int vv : adj[u]) {
      deg[vv]--;
      if (deg[vv] == k - 1) q.push(vv);
    }
  }
  vector<int> vals;
  for (int i = 1; i <= n; i++) {
    if (!marked[i]) vals.push_back(i);
  }
  if (vals.size() != 0) {
    cout << 1 << " " << vals.size() << endl;
    for (int vv : vals) cout << vv << " ";
    cout << endl;
    return;
  }
  if (k >= 500) {
    cout << -1 << endl;
    return;
  }
  for (int i = 1; i <= n; i++) {
    deg[i] = odeg[i];
    marked[i] = false;
    if (deg[i] < k - 1) q.push(i);
  }
  while (q.size()) {
    u = q.front();
    q.pop();
    marked[u] = true;
    for (int vv : adj[u]) {
      deg[vv]--;
      if (deg[vv] == k - 2) q.push(vv);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (deg[i] == k - 1) q.push(i);
  }
  while (q.size()) {
    u = q.front();
    q.pop();
    if (deg[u] == k - 1) {
      vector<int> cstuff;
      cstuff.push_back(u);
      for (int vv : adj[u]) {
        if (!marked[vv]) {
          cstuff.push_back(vv);
        }
      }
      bool isok = true;
      for (int i = 0; i < cstuff.size(); i++) {
        if (!isok) break;
        for (int j = i + 1; j < cstuff.size(); j++) {
          if (ed[cstuff[i]].count(cstuff[j]) == 0) {
            isok = false;
            break;
          }
        }
      }
      if (isok) {
        cout << 2 << endl;
        for (int vv : cstuff) cout << vv << " ";
        cout << endl;
        return;
      }
    }
    marked[u] = true;
    for (int vv : adj[u]) {
      deg[vv]--;
      if (deg[vv] == k - 1) q.push(vv);
    }
  }
  cout << -1 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> t;
  while (t--) {
    solve();
  }
}