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 std::max;
using std::min;
const int inf = 0x3f3f3f3f, Inf = 0x7fffffff;
const long long INF = 0x3f3f3f3f3f3f3f3f;
std::mt19937 rnd(std::chrono::steady_clock::now().time_since_epoch().count());
template <typename _Tp>
_Tp gcd(const _Tp &a, const _Tp &b) {
  return (!b) ? a : gcd(b, a % b);
}
template <typename _Tp>
inline _Tp abs(const _Tp &a) {
  return a >= 0 ? a : -a;
}
template <typename _Tp>
inline void chmax(_Tp &a, const _Tp &b) {
  (a < b) && (a = b);
}
template <typename _Tp>
inline void chmin(_Tp &a, const _Tp &b) {
  (b < a) && (a = b);
}
template <typename _Tp>
inline void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (!isdigit(ch)) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (isdigit(ch)) x = (((x << 2) + x) << 1) + (ch & 15), ch = getchar();
  f && (x = -x);
}
template <typename _Tp, typename... Args>
inline void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
inline int read_str(char *s) {
  char ch(getchar());
  while (ch == ' ' || ch == '\r' || ch == '\n') ch = getchar();
  char *tar = s;
  *tar = ch, ch = getchar();
  while (ch != ' ' && ch != '\r' && ch != '\n' && ch != EOF)
    *(++tar) = ch, ch = getchar();
  return tar - s + 1;
}
const int N = 100005;
std::vector<int> e[N];
int dg[N];
struct node {
  int u, d;
  inline bool operator<(const node &o) const { return d > o.d; }
};
std::priority_queue<node> q;
bool mark[N], used[N];
int ans[N];
std::vector<int> Q[N];
void MAIN() {
  int n, m, k;
  read(n, m, k);
  if (k > 450) return (void)puts("-1");
  for (int i = 1; i <= n; ++i) e[i].clear(), dg[i] = 0;
  int x, y;
  for (int i = 1; i <= m; ++i)
    read(x, y), e[x].push_back(y), e[y].push_back(x), ++dg[x], ++dg[y];
  for (int i = 0; i <= n; ++i) used[i] = mark[i] = false, Q[i].clear();
  while (!q.empty()) q.pop();
  for (int i = 1; i <= n; ++i) q.push({i, dg[i]});
  std::vector<std::vector<int>> clique;
  while (!q.empty()) {
    node x = q.top();
    q.pop();
    if (dg[x.u] != x.d || used[x.u]) continue;
    if (dg[x.u] == k - 1) {
      std::vector<int> v = {x.u};
      for (auto it : e[x.u])
        if (!used[it]) v.push_back(it);
      clique.push_back(v);
    }
    if (dg[x.u] >= k) {
      std::vector<int> v;
      for (int i = 1; i <= n; ++i)
        if (dg[i] >= k) v.push_back(i);
      printf("1 %d\n", ((int)v.size()));
      for (auto it : v) printf("%d ", it);
      printf("\n");
      return;
    }
    used[x.u] = true;
    for (auto v : e[x.u]) {
      --dg[v];
      if (!used[v]) q.push({v, dg[v]});
    }
  }
  for (int i = 0; i < ((int)clique.size()); ++i) ans[i] = 0;
  for (int id = 0; id < ((int)clique.size()); ++id)
    for (int i = 0; i < ((int)clique[id].size()); ++i)
      Q[clique[id][i]].push_back(id);
  for (int i = 1; i <= n; ++i) {
    for (auto v : e[i]) mark[v] = true;
    for (auto it : Q[i])
      for (auto v : clique[it]) ans[it] += mark[v];
    for (auto v : e[i]) mark[v] = false;
  }
  for (int i = 0; i < ((int)clique.size()); ++i) {
    if (ans[i] == k * (k - 1)) {
      printf("2\n");
      for (auto it : clique[i]) printf("%d ", it);
      printf("\n");
      return;
    }
  }
  puts("-1");
}
int main() {
  int _;
  read(_);
  while (_--) MAIN();
  return 0;
}