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 dadsadasda;
const int INF = 0x3f3f3f3f;
const long long LLINF = 1e18;
const int MAXN = 1e5 + 10;
int n, m, k;
vector<int> G[MAXN];
int deg[MAXN];
bool erased[MAXN];
queue<int> pos_cliq, pos_set;
void erase(int v) {
  erased[v] = true;
  for (auto u : G[v]) {
    if (erased[u]) continue;
    m--;
    deg[u]--;
    if (deg[u] < k - 1) erase(u);
  }
}
void erase2(int v) {
  erased[v] = true;
  for (auto u : G[v]) {
    if (erased[u]) continue;
    deg[u]--;
    if (deg[u] < k - 1) erase2(u);
    if (deg[u] == k - 1) pos_cliq.push(u);
  }
}
bool check_clique(int a) {
  long long aux = k;
  aux = (aux * (aux - 1)) >> 1;
  if (aux > (long long)m) return false;
  for (auto b : G[a]) {
    if (erased[b]) continue;
    int found = 0;
    for (auto c : G[a]) {
      if (erased[c]) continue;
      auto it = lower_bound(G[b].begin(), G[b].end(), c);
      if (it == G[b].end()) continue;
      if (*it == c) found++;
    }
    if (found < k - 2) return false;
  }
  printf("2\n");
  printf("%d ", a + 1);
  for (auto b : G[a]) {
    if (erased[b]) continue;
    printf("%d ", b + 1);
  }
  printf("\n");
  return true;
}
void solve() {
  dadsadasda = scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    int a, b;
    dadsadasda = scanf("%d %d", &a, &b);
    a--, b--;
    G[a].push_back(b);
    G[b].push_back(a);
    deg[a]++, deg[b]++;
  }
  long long aux = k;
  aux = (aux * (aux - 1)) >> 1;
  if (aux > m) {
    printf("-1\n");
    return;
  }
  for (int i = 0; i < n; i++) {
    if (!erased[i] && deg[i] < k - 1) erase(i);
  }
  for (int i = 0; i < n; i++) {
    sort(G[i].begin(), G[i].end());
    if (deg[i] == k - 1)
      pos_cliq.push(i);
    else
      pos_set.push(i);
  }
  while (!pos_cliq.empty()) {
    int v = pos_cliq.front();
    pos_cliq.pop();
    if (erased[v]) continue;
    if (check_clique(v))
      return;
    else
      erase2(v);
  }
  while (!pos_set.empty()) {
    int v = pos_set.front();
    pos_set.pop();
    if (erased[v]) continue;
    vector<int> ans;
    for (int i = 0; i < n; i++) {
      if (!erased[i]) ans.push_back(i);
    }
    printf("%d %d\n", 1, (int)ans.size());
    for (auto x : ans) {
      printf("%d ", x + 1);
    }
    printf("\n");
    return;
  }
  printf("-1\n");
}
int main() {
  int t;
  dadsadasda = scanf("%d", &t);
  while (t--) {
    solve();
    for (int i = 0; i < n; i++) {
      erased[i] = false;
      G[i].clear();
      deg[i] = 0;
      while (!pos_set.empty()) pos_set.pop();
      while (!pos_cliq.empty()) pos_cliq.pop();
    }
  }
  return 0;
}