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;
const int maxN = 201000;
vector<int> g[maxN];
set<int> gs[maxN];
bool cmp(int x, int y) { return gs[x].size() < gs[y].size(); }
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < n; ++i) {
    g[i].clear();
    gs[i].clear();
  }
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    g[x].push_back(y);
    g[y].push_back(x);
    gs[x].insert(y);
    gs[y].insert(x);
  }
  set<pair<int, int> > degs;
  for (int i = 0; i < n; ++i) {
    degs.insert(make_pair(gs[i].size(), i));
  }
  while (!degs.empty()) {
    auto cur = *degs.begin();
    if (cur.first >= k) {
      cout << "1"
           << " " << degs.size() << endl;
      for (auto v : degs) {
        cout << v.second + 1 << " ";
      }
      cout << endl;
      return;
    }
    degs.erase(degs.begin());
    int v = cur.second;
    for (auto to : gs[v]) {
      degs.erase(make_pair(gs[to].size(), to));
      gs[to].erase(v);
      degs.insert(make_pair(gs[to].size(), to));
    }
  }
  for (int i = 0; i < n; ++i) {
    gs[i].clear();
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < g[i].size(); ++j) {
      gs[i].insert(g[i][j]);
      gs[g[i][j]].insert(i);
    }
  }
  degs.clear();
  for (int i = 0; i < n; ++i) {
    degs.insert(make_pair(gs[i].size(), i));
  }
  while (!degs.empty()) {
    auto cur = *degs.begin();
    if (cur.first >= k - 1) {
      break;
    }
    degs.erase(degs.begin());
    int v = cur.second;
    for (auto to : gs[v]) {
      degs.erase(make_pair(gs[to].size(), to));
      gs[to].erase(v);
      degs.insert(make_pair(gs[to].size(), to));
    }
  }
  if (k >= 500) {
    cout << "-1" << endl;
    return;
  }
  vector<int> used(n, 0);
  while (!degs.empty()) {
    auto v = *degs.begin();
    if (v.first >= k) {
      break;
    }
    int cv = v.second;
    bool should_start = !used[cv];
    vector<int> neighbors;
    for (auto to : gs[cv]) {
      neighbors.push_back(to);
    }
    neighbors.push_back(cv);
    for (int i = 0; i < neighbors.size(); ++i) {
      if (gs[neighbors[i]].size() == k - 1) {
        used[neighbors[i]] = 1;
      }
    }
    sort(neighbors.begin(), neighbors.end(), cmp);
    if (should_start) {
      bool good = true;
      for (int i = 0; i < neighbors.size() && good; ++i) {
        for (int j = 0; j < neighbors.size() && good; ++j) {
          if (i == j) {
            continue;
          }
          int v1 = neighbors[i], v2 = neighbors[j];
          if (!gs[v1].count(v2)) {
            good = false;
            break;
          }
        }
      }
      if (good) {
        cout << 2 << endl;
        for (int i = 0; i < neighbors.size(); ++i) {
          cout << neighbors[i] + 1 << " ";
        }
        cout << endl;
        return;
      }
    }
    degs.erase(degs.begin());
    for (auto to : gs[cv]) {
      degs.erase(make_pair(gs[to].size(), to));
      gs[to].erase(cv);
      degs.insert(make_pair(gs[to].size(), to));
    }
  }
  cout << -1 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  for (int tt = 0; tt < t; ++tt) {
    solve();
  }
  return 0;
}