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;
char _buf[100000], *_p1 = _buf, *_p2 = _buf;
inline int gi() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return (f == 1) ? x : -x;
}
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
const int maxn = 2e5 + 5;
int n, m, T;
vector<int> vec[maxn];
int ind[maxn], k;
set<int> s[maxn];
bool vis[maxn], vis2[maxn];
vector<int> ans;
queue<int> q;
inline void init() {
  for (int i = 1; i <= n; ++i)
    vec[i].clear(), ind[i] = 0, s[i].clear(), vis[i] = 0, vis2[i] = 0;
  ans.clear();
  while (!q.empty()) q.pop();
}
inline void input() {
  n = gi(), m = gi(), k = gi();
  for (int i = 1; i <= m; ++i) {
    int u = gi(), v = gi();
    vec[u].push_back(v), vec[v].push_back(u);
    ind[u]++, ind[v]++;
  }
}
inline void dfs(int u) {
  vis2[u] = 1;
  ans.push_back(u);
  for (auto v : vec[u]) {
    if (!vis[v] && !vis2[v]) {
      dfs(v);
    }
  }
}
inline bool check(int u, int xx) {
  vector<int> vec1;
  vec1.clear();
  for (auto x : s[u]) {
    if (ind[x] < k - 1) return 0;
    for (auto y : vec1) {
      if (!s[x].count(y)) {
        return 0;
      }
    }
    vec1.push_back(x);
  }
  return 1;
}
inline void solve(int xx) {
  if (k == 1) {
    printf("2\n1\n");
    return;
  }
  for (int i = 1; i <= n; ++i) {
    for (auto x : vec[i]) s[i].insert(x);
    ind[i] = s[i].size();
  }
  for (int i = 1; i <= n; ++i) {
    if (ind[i] < k) q.push(i), vis[i] = 1;
    if (ind[i] == k - 1) {
      if (1LL * k * (k - 1) <= 2 * m && check(i, xx)) {
        printf("2\n");
        for (auto x : s[i]) printf("%d ", x);
        printf("%d\n", i);
        return;
      }
    }
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (auto v : s[u]) {
      ind[v]--;
      s[v].erase(u);
      if (!vis[v] && ind[v] < k) vis[v] = 1, q.push(v);
      if (1LL * k * k <= 2 * m && ind[v] == k - 1) {
        if (check(v, xx)) {
          printf("2\n");
          for (auto x : s[v]) printf("%d ", x);
          printf("%d\n", v);
          return;
        }
      }
    }
    s[u].clear();
  }
  bool ok = 0;
  for (int i = 1; i <= n; ++i) {
    if (!vis[i]) {
      ok = 1;
      dfs(i);
      break;
    }
  }
  if (ok) {
    printf("1 %d\n", (int)(ans.size()));
    for (auto x : ans) printf("%d ", x);
    printf("\n");
    return;
  }
  printf("-1\n");
}
int main() {
  T = gi();
  for (int i = 1; i <= T; ++i) {
    init();
    input();
    solve(i);
  }
  return 0;
}