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;
const int maxn = 3e5 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f, B = 133597;
struct HT {
  const static int M = 29847577;
  bool a[M];
  int buf[maxn], tot;
  inline void ins(long long x) { a[buf[++tot] = x % M] = 1; }
  inline bool find(long long x) { return a[x % M]; }
  inline void clear() {
    for (int i = 1; i <= tot; ++i) a[buf[i]] = 0;
    tot = 0;
  }
} ht;
inline long long Hash(long long x, long long y) {
  if (x > y) swap(x, y);
  return x * B + y;
}
vector<int> e[maxn];
int deg[maxn], inq[maxn], vis[maxn], buf[maxn], tot;
vector<int> ans;
inline bool check(int u, int k) {
  buf[tot = 1] = u;
  for (auto v : e[u]) {
    if (vis[v]) continue;
    for (int i = 1; i <= tot; ++i)
      if (!ht.find(Hash(buf[i], v))) return 0;
    buf[++tot] = v;
  }
  if (tot < k) return 0;
  for (int i = 1; i <= tot; ++i) ans.push_back(buf[i]);
  return 1;
}
inline void tupo(int n, int k) {
  queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (deg[i] < k) q.push(i), vis[i] = 1;
  while (int(q.size())) {
    int u = q.front();
    q.pop();
    for (auto v : e[u]) {
      deg[v]--;
      if (!vis[v] && deg[v] < k) q.push(v), vis[v] = 1;
    }
  }
}
int sol(int n, int k) {
  tupo(n, k - 1);
  queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (!vis[i] && deg[i] < k) q.push(i), inq[i] = 1;
  while (int(q.size())) {
    int u = q.front();
    q.pop();
    if (check(u, k)) return 2;
    vis[u] = 1;
    for (auto v : e[u]) {
      deg[v]--;
      if (!vis[v] && !inq[v] && deg[v] < k) q.push(v), inq[v] = 1;
    }
  }
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) ans.push_back(i);
  return int(ans.size()) ? 1 : 0;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    ans.clear();
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; ++i) vis[i] = deg[i] = inq[i] = 0, e[i].clear();
    for (int i = 1; i <= m; ++i) {
      int u, v;
      scanf("%d%d", &u, &v);
      e[u].push_back(v);
      e[v].push_back(u);
      deg[u]++;
      deg[v]++;
      ht.ins(Hash(u, v));
    }
    int res = sol(n, k);
    if (!res)
      printf("-1\n");
    else {
      if (res == 1)
        printf("1 %d\n", int(ans.size()));
      else
        printf("2\n");
      for (int i = 0; i < int(ans.size()); ++i)
        printf("%d%c", ans[i], " \n"[i == int(ans.size()) - 1]);
    }
    ht.clear();
  }
  return 0;
}