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;
void Start() { ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0); }
bool endline = false;
bool endlineA = false;
template <class T>
istream& operator>>(istream& in, vector<T>& v) {
  for (auto& i : v) {
    in >> i;
  }
  return in;
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  for (auto i : v) {
    out << i;
    if (!endline) {
      out << ' ';
    } else {
      out << '\n';
    }
  }
  if (!endline && endlineA) {
    out << '\n';
  }
  return out;
}
template <class T, class U>
istream& operator>>(istream& in, pair<T, U>& p) {
  in >> p.first >> p.second;
  return in;
}
template <class T, class U>
ostream& operator<<(ostream& out, const pair<T, U>& p) {
  out << p.first << ' ' << p.second;
  return out;
}
long long gettime() {
  return chrono::high_resolution_clock::now().time_since_epoch().count() /
         1000000;
}
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long randnum(long long l, long long r) { return l + rnd() % (r - l + 1); }
signed main() {
  Start();
  int q;
  cin >> q;
  while (q--) {
    int n, m, k;
    cin >> n >> m >> k;
    vector<unordered_set<int>> g(n);
    for (int i = 0; i < (m); i++) {
      int a, b;
      cin >> a >> b;
      a--;
      b--;
      g[a].insert(b);
      g[b].insert(a);
    }
    if ((long long)k * (long long)(k - 1) / 2 > m) {
      cout << "-1\n";
      continue;
    }
    auto start = g;
    queue<int> q;
    vector<bool> used(n, false);
    for (int i = 0; i < (n); i++)
      if (g[i].size() < k) q.push(i), used[i] = true;
    while (!q.empty()) {
      int v = q.front();
      q.pop();
      while (!g[v].empty()) {
        int u = *g[v].begin();
        g[v].erase(g[v].begin());
        g[u].erase(v);
        if (g[u].size() < k && !used[u]) {
          used[u] = true;
          q.push(u);
        }
      }
    }
    vector<int> res1;
    for (int i = 0; i < (n); i++) {
      if (!used[i]) {
        res1.push_back(i);
      }
    }
    if (!res1.empty()) {
      cout << "1 " << res1.size() << '\n';
      for (auto i : res1) cout << i + 1 << ' ';
      cout << '\n';
      continue;
    }
    g = start;
    used.assign(n, false);
    for (int i = 0; i < (n); i++)
      if (g[i].size() < k - 1) q.push(i), used[i] = true;
    while (!q.empty()) {
      int v = q.front();
      q.pop();
      while (!g[v].empty()) {
        int u = *g[v].begin();
        g[v].erase(g[v].begin());
        g[u].erase(v);
        if (g[u].size() < k - 1 && !used[u]) {
          used[u] = true;
          q.push(u);
        }
      }
    }
    for (int i = 0; i < (n); i++) {
      if (!used[i] && g[i].size() == k - 1) {
        q.push(i);
      }
    }
    bool done = false;
    while (!q.empty()) {
      int v = q.front();
      q.pop();
      vector<int> c;
      c.push_back(v);
      for (auto u : g[v]) {
        c.push_back(u);
      }
      bool good = true;
      for (auto i : c) {
        int sum = 1;
        for (auto u : c) sum += (g[i].find(u) != g[i].end());
        if (sum != k) {
          good = false;
          break;
        }
      }
      if (good) {
        done = true;
        cout << "2\n";
        for (auto i : c) cout << i + 1 << ' ';
        cout << '\n';
        break;
      }
      while (!g[v].empty()) {
        int u = *g[v].begin();
        g[v].erase(g[v].begin());
        g[u].erase(v);
        if (g[u].size() == k - 1 && !used[u]) {
          used[u] = true;
          q.push(u);
        }
      }
    }
    if (!done) {
      cout << "-1\n";
    }
  }
  return 0;
}