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 = 100005;
int n, m, k, deg[maxn], e[maxn], ok[maxn], id[maxn], vis[maxn], tot;
vector<int> g[maxn];
set<int> s[maxn];
inline void toposort(int k) {
  for (int i = 1; i <= n; i++) ok[i] = 1, e[i] = deg[i];
  queue<int> q;
  while (!q.empty()) q.pop();
  for (int i = 1; i <= n; i++)
    if (e[i] < k) ok[i] = 0, q.push(i);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (auto v : g[u])
      if (ok[v]) {
        e[v]--;
        if (e[v] < k) ok[v] = 0, q.push(v);
      }
  }
}
void dfs(int x) {
  ok[x] = 0, id[++tot] = x;
  for (auto y : g[x])
    if (ok[y]) dfs(y);
}
inline void solve() {
  scanf("%d%d%d", &n, &m, &k);
  int x, y;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
    deg[x]++, deg[y]++;
  }
  if (k == 1) {
    printf("2\n1\n");
    return;
  }
  if ((long long)k * (k - 1) > 2 * m) {
    puts("-1");
    return;
  }
  toposort(k);
  for (int i = 1; i <= n; i++)
    if (ok[i]) {
      tot = 0, dfs(i);
      printf("1 %d\n", tot);
      for (int i = 1; i <= tot; i++) printf("%d ", id[i]);
      putchar('\n');
      return;
    }
  toposort(k - 1);
  for (int i = 1; i <= n; i++)
    if (ok[i])
      for (auto j : g[i])
        if (ok[j]) s[i].insert(j);
  queue<int> q;
  while (!q.empty()) q.pop();
  for (int i = 1; i <= n; i++)
    if (ok[i] && e[i] == k - 1) q.push(i);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (e[u] == k - 1) {
      id[tot = 1] = u;
      for (auto v : g[u])
        if (ok[v]) id[++tot] = v;
      int fl = 1;
      for (int i = 2; i <= tot && fl; i++)
        for (int j = i + 1; j <= tot && fl; j++)
          if (s[id[i]].find(id[j]) == s[id[i]].end()) fl = 0;
      if (fl) {
        printf("2\n");
        for (int i = 1; i <= tot; i++) printf("%d ", id[i]);
        putchar('\n');
        return;
      }
    }
    ok[u] = 0;
    for (auto v : g[u])
      if (ok[v] && --e[v] == k - 1) q.push(v);
  }
  puts("-1");
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    solve();
    for (int i = 1; i <= n; i++) g[i].clear(), s[i].clear(), deg[i] = 0;
  }
  return 0;
}