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 <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return {i, j};
}
struct printer {
  ~printer() { cerr << endl; }
  template <class c>
  printer& operator,(c x) {
    cerr << boolalpha << x;
    return *this;
  }
  printer& operator,(string x) {
    cerr << x;
    return *this;
  }
  template <class c, class d>
  printer& operator,(pair<c, d> x) {
    return *this, "(", x.first, ", ", x.second, ")";
  }
  template <class... d, template <class...> class c>
  printer& operator,(c<d...> x) {
    return *this, range(begin(x), end(x));
  }
  template <class c>
  printer& operator,(rge<c> x) {
    *this, "[";
    for (auto it = x.b; it != x.e; ++it) *this, (it == x.b ? "" : ", "), *it;
    return *this, "]";
  }
};
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int my_rand(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
const int magic = 40;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    int n, m, K;
    cin >> n >> m >> K;
    vector<vector<int>> g(n + 1);
    vector<int> mapped(n + 1, -1), rev(n + 1, -1);
    int sz = 0;
    for (int i = 1; i <= m; i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    vector<int> deg(n + 1);
    set<pair<int, int>> pq;
    for (int i = 1; i <= n; i++) {
      sort(g[i].begin(), g[i].end());
      deg[i] = g[i].size();
      pq.insert({deg[i], i});
      if (deg[i] >= K - 1) {
        mapped[i] = sz;
        rev[sz] = i;
        sz++;
      }
    }
    vector<bitset<5000>> edge;
    if (K > magic) {
      bitset<5000> e;
      for (int i = 1; i <= n; i++) {
        if (mapped[i] != -1) {
          bitset<5000> e;
          for (int v : g[i]) {
            if (mapped[v] != -1) {
              e[mapped[v]] = 1;
            }
          }
          e[mapped[i]] = 1;
          edge.push_back(e);
        }
      }
    }
    int ans = -1;
    vector<bool> done(n + 1);
    vector<int> clique;
    while (pq.size() > 0) {
      auto t = *pq.begin();
      int u = t.second;
      assert(deg[u] == t.first);
      if (deg[u] >= K) {
        ans = 1;
        break;
      }
      pq.erase(t);
      done[u] = true;
      for (int v : g[u]) {
        if (done[v]) continue;
        pq.erase({deg[v], v});
        deg[v]--;
        pq.insert({deg[v], v});
      }
      if (deg[u] < K - 1) continue;
      vector<int> nei;
      for (int v : g[u]) {
        if (done[v] == false) {
          nei.push_back(v);
        }
      }
      assert(nei.size() == K - 1);
      bool found = true;
      if (K <= magic) {
        for (int v : nei) {
          for (int w : nei) {
            if (v == w) continue;
            int lo = 0, hi = (int)g[v].size() - 1;
            while (hi > lo) {
              int mid = (lo + hi + 1) / 2;
              if (g[v][mid] <= w)
                lo = mid;
              else
                hi = mid - 1;
            }
            if (g[v][lo] != w) {
              found = false;
              break;
            }
          }
          if (found == false) break;
        }
      } else {
        for (int v : nei) {
          if (mapped[v] == -1) {
            found = false;
            break;
          }
        }
        if (found) {
          bitset<5000> e;
          for (int x : nei) {
            e[mapped[x]] = 1;
          }
          e[mapped[u]] = 1;
          for (int v : nei) {
            if ((e & edge[mapped[v]]).count() != K) {
              found = false;
              break;
            }
          }
        }
      }
      if (found == true) {
        ans = 2;
        clique = nei;
        clique.push_back(u);
        break;
      }
    }
    if (ans == 1) {
      vector<int> res;
      for (int i = 1; i <= n; i++)
        if (done[i] == false) res.push_back(i);
      cout << 1 << " " << res.size() << "\n";
      for (int x : res) cout << x << " ";
      cout << "\n";
    } else if (ans == -1) {
      cout << "-1"
           << "\n";
    } else {
      cout << "2"
           << "\n";
      for (int x : clique) cout << x << " ";
      cout << "\n";
    }
  }
}