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 T>
using vc = vector<T>;
template <class T>
using vvc = vc<vc<T>>;
template <class T>
void mkuni(vector<T> &v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
long long rand_int(long long l, long long r) {
  static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
  return uniform_int_distribution<long long>(l, r)(gen);
}
template <class T>
void print(T x, int suc = 1) {
  cout << x;
  if (suc == 1)
    cout << '\n';
  else
    cout << ' ';
}
template <class T>
void print(const vector<T> &v, int suc = 1) {
  for (int i = 0; i < v.size(); i++)
    print(v[i], i == (int)(v.size()) - 1 ? suc : 2);
}
struct D {
  mutable int d;
  int id;
  bool operator<(const D &a) const {
    if (d == a.d) return id < a.id;
    return d > a.d;
  }
};
const int N = 2e5 + 7;
template <typename T1, typename T2>
struct Table {
  T1 key[N];
  T2 val[N];
  int head[19260817], next[N], tot;
  int sta[N], top;
  void init(bool flag) {
    if (flag) {
      memset(head, -1, sizeof(head));
      tot = top = 0;
      return;
    }
    while (top > 0) {
      head[sta[--top]] = -1;
    }
    tot = 0;
  }
  int gethas(T1 v) { return (v.first * 1000 + v.second) % 19260817; }
  void modify(T1 x, T2 y) {
    int u = gethas(x);
    for (int i = head[u]; ~i; i = next[i]) {
      if (key[i] == x) {
        val[i] = y;
        return;
      }
    }
    key[tot] = x;
    val[tot] = y;
    next[tot] = head[u];
    if (head[u] == -1) {
      sta[top++] = u;
    }
    head[u] = tot++;
  }
  T2 query(T1 x) {
    int u = gethas(x);
    for (int i = head[u]; ~i; i = next[i]) {
      if (key[i] == x) {
        return val[i];
      }
    }
    return 0;
  }
};
Table<pair<int, int>, int> es;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  es.init(1);
  while (t--) {
    es.init(0);
    bool fl = 0;
    int n, m, k;
    cin >> n >> m >> k;
    vector<int> deg(n + 1);
    vc<vector<int>> g(n + 1);
    for (int i = 0; i < m; ++i) {
      int u, v;
      cin >> u >> v;
      g[u].emplace_back(v);
      g[v].emplace_back(u);
      es.modify(pair<int, int>(u, v), 1);
      es.modify(pair<int, int>(v, u), 1);
    }
    for (int i = 1; i <= n; ++i) {
      deg[i] = g[i].size();
    }
    priority_queue<D> q;
    vector<int> vis(n + 1, 0);
    for (int i = 1; i <= n; ++i) {
      if (deg[i] < k) q.push(D{deg[i], i}), vis[i] = 1;
    }
    vector<int> clique;
    while (q.size()) {
      auto now = q.top();
      int cur = now.id;
      ;
      q.pop();
      vis[cur] = -1;
      if (deg[cur] == k - 1 && 1ll * k * (k - 1) / 2 <= m) {
        clique.emplace_back(cur);
        bool suc = 1;
        for (auto &u : g[cur]) {
          if (vis[u] != -1) clique.emplace_back(u);
        }
        for (auto &u : clique) {
          for (auto &v : clique) {
            if (u == v) continue;
            if (!es.query(pair<int, int>(u, v))) suc = 0;
          }
          if (!suc) break;
        };
        if (!suc)
          clique.clear();
        else {
          cout << "2\n";
          print(clique);
          fl = 1;
          break;
        }
      }
      for (auto &u : g[cur]) {
        if (--deg[u] < k && vis[u] == 0) q.push(D{deg[u], u}), vis[u] = 1;
      }
    }
    if (!fl) {
      bool suc = 0;
      vector<int> ans;
      for (int i = 1; i <= n; ++i) {
        if (vis[i] == 0) suc = 1, ans.emplace_back(i);
      }
      if (suc) {
        cout << 1 << " " << ans.size() << '\n';
        print(ans);
      } else
        cout << "-1\n";
    }
  }
}