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;
int gi() {
  char cc = getchar();
  int cn = 0, flus = 1;
  while (cc < '0' || cc > '9') {
    if (cc == '-') flus = -flus;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') cn = cn * 10 + cc - '0', cc = getchar();
  return cn * flus;
}
const int N = 2e5 + 5;
const int P = 2333333;
int n, m, K, Ty, deg[N], vis[N], ST[N], num;
vector<int> G[N];
priority_queue<pair<int, int>> q;
struct hashmap {
  static const int N = 1e6 + 61, P = 999983;
  struct node {
    int next;
    long long key;
  } e[N];
  int head[N], cnt, st[N], top;
  inline void ins(long long x) {
    int hsh = 1ll * x % P + 1;
    for (int i = head[hsh]; i; i = e[i].next)
      if (e[i].key == x) return;
    e[++cnt] = (node){head[hsh], x}, head[hsh] = cnt;
    st[++top] = hsh;
  }
  inline void clear() {
    cnt = 0;
    for (register int i = (1); i <= (top); ++i) head[st[i]] = 0;
    top = 0;
  }
  inline int query(long long x) {
    int hsh = 1ll * x % P + 1;
    for (int i = head[hsh]; i; i = e[i].next)
      if (e[i].key == x) return 1;
    return 0;
  }
} tab;
vector<int> ans;
void Dfs(int x) {
  vis[x] = 1, ans.push_back(x);
  if (Ty)
    for (int v : G[x])
      if (!vis[v]) printf("fuck go this %d -> %d\n", x, v);
  for (int v : G[x])
    if (!vis[v]) Dfs(v);
}
void solve() {
  n = gi(), m = gi(), K = gi();
  if (K > 450) {
    puts("-1");
    exit(0);
  }
  int x, y, type = 0;
  for (register int i = (1); i <= (m); ++i)
    x = gi(), y = gi(), G[x].push_back(y), G[y].push_back(x), ++deg[x],
    ++deg[y], tab.ins(1ll * x * n + y), tab.ins(1ll * y * n + x);
  for (register int i = (1); i <= (n); ++i) q.push(make_pair(-deg[i], i));
  int ctt = 0;
  while (!q.empty()) {
    if (!ans.empty()) break;
    pair<int, int> u = q.top();
    q.pop();
    if (vis[u.second]) continue;
    int D = -u.first;
    if (D == K - 1) {
      num = 0;
      ST[++num] = u.second;
      for (int v : G[u.second])
        if (!vis[v]) ST[++num] = v;
      int flag = 1;
      ctt += K * K;
      for (register int i = 1; i < num; ++i)
        for (register int j = i + 1; j <= num; ++j) {
          flag &= tab.query(1ll * ST[i] * n + ST[j]);
        }
      if (flag) {
        for (register int j = (1); j <= (num); ++j) ans.push_back(ST[j]);
        type = 2;
        break;
      }
    } else if (D >= K) {
      type = 1, Dfs(u.second);
      break;
    }
    vis[u.second] = 1;
    for (int v : G[u.second]) {
      --deg[v], q.push(make_pair(-deg[v], v));
    }
    if (ctt >= 2e7) {
      printf("%d\n", ctt);
      exit(0);
    }
  }
  if (!type)
    puts("-1");
  else {
    if (type == 1)
      printf("1 %d\n", (int)ans.size());
    else
      puts("2");
    for (int u : ans) printf("%d ", u);
    puts("");
  }
  for (register int i = (1); i <= (n); ++i) G[i].clear(), vis[i] = deg[i] = 0;
  ans.clear();
  tab.clear();
  while (!q.empty()) q.pop();
}
signed main() {
  int T = gi();
  while (T--) solve();
  return 0;
}