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 INF = 0x3f3f3f3f;
const long long INFLL = 1e18;
const int MOD = 1e9 + 7;
const int MAXN = 2e5 + 5;
vector<int> G[MAXN];
int deg[MAXN];
int n, m, k;
queue<int> toRem;
queue<int> clique;
bool ans[MAXN];
int remaining;
bool exists(int u, int v) {
  auto it = lower_bound(G[u].begin(), G[u].end(), v);
  return it != G[u].end() && *it == v;
}
bool check(int v) {
  for (int x : G[v])
    if (ans[x])
      for (int y : G[v])
        if (ans[y] && x > y)
          if (!exists(x, y)) return false;
  return true;
}
void remove(int v) {
  if (!ans[v]) return;
  ans[v] = false;
  remaining--;
  for (int u : G[v]) {
    if (!ans[u]) continue;
    deg[u]--;
    if (deg[u] == k - 2) toRem.push(u);
    if (deg[u] == k - 1) clique.push(u);
  }
}
void solve() {
  for (int i = 0; i < (int)n; i++) sort(G[i].begin(), G[i].end());
  for (int i = 0; i < (int)n; i++) deg[i] = G[i].size();
  for (int i = 0; i < (int)n; i++) {
    if (deg[i] < k - 1) toRem.push(i);
    if (deg[i] == k - 1) clique.push(i);
  }
  for (int i = 0; i < (int)n; i++) ans[i] = true;
  remaining = n;
  while (!toRem.empty() || !clique.empty()) {
    while (!toRem.empty()) {
      remove(toRem.front());
      toRem.pop();
    }
    while (!clique.empty() && !ans[clique.front()]) {
      clique.pop();
    }
    if (!clique.empty()) {
      int v = clique.front();
      clique.pop();
      assert(deg[v] == k - 1);
      if (check(v)) {
        printf("2\n");
        printf("%d", v + 1);
        for (int u : G[v])
          if (ans[u]) printf(" %d", u + 1);
        printf("\n");
        return;
      } else {
        remove(v);
      }
    }
  }
  if (remaining == 0)
    printf("-1\n");
  else {
    printf("1 %d\n", remaining);
    for (int i = 0; i < (int)n; i++)
      if (ans[i]) printf("%d ", i + 1);
    printf("\n");
  }
}
void clearAll() {
  for (int i = 0; i < (int)n; i++) G[i].clear();
  while (!clique.empty()) clique.pop();
  while (!toRem.empty()) toRem.pop();
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n), scanf("%d", &m), scanf("%d", &k);
    for (int i = 0; i < (int)m; i++) {
      int u, v;
      scanf("%d", &u), scanf("%d", &v);
      u--, v--;
      G[u].push_back(v);
      G[v].push_back(u);
    }
    solve();
    clearAll();
  }
  return 0;
}