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;
unordered_set<int> e[200005];
int deg[200005], deg1[200005];
bool use[200005];
struct Item {
  int u, d;
  bool operator<(const Item& it) const { return d > it.d; }
};
int solve(int k, int n) {
  priority_queue<Item> q;
  for (int i = 1; i <= n; i++) {
    use[i] = true;
    if (deg[i] < k) {
      q.push(Item{i, deg[i]});
      use[i] = false;
    }
  }
  while (!q.empty()) {
    Item it = q.top();
    q.pop();
    for (int v : e[it.u]) {
      deg[v]--;
      deg[it.u]--;
      if (deg[v] == k - 1) {
        use[v] = false;
        q.push(Item{v, deg[v]});
      }
    }
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (use[i]) cnt++;
  return cnt;
}
bool check(int x) {
  int cnt = 0;
  unordered_set<int> candidates;
  for (int v : e[x]) {
    if (use[v]) {
      candidates.insert(v);
    }
  }
  candidates.insert(x);
  if (candidates.size() > 500)
    return false;
  else {
    for (int v : candidates) {
      for (int v1 : candidates) {
        if (v == v1) continue;
        if (e[v].find(v1) == e[v].end()) return false;
      }
    }
  }
  return true;
}
int main(void) {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, m, k;
    scanf("%d %d %d", &n, &m, &k);
    for (int i = 1; i <= n; i++) deg[i] = 0;
    for (int i = 0; i < m; i++) {
      int u, v;
      scanf("%d %d", &u, &v);
      e[u].insert(v);
      e[v].insert(u);
      deg[u]++;
      deg[v]++;
    }
    int ans = 0;
    for (int i = 1; i <= n; i++) deg1[i] = deg[i];
    if ((ans = solve(k, n))) {
      printf("1 %d\n", ans);
      for (int i = 1; i <= n; i++)
        if (use[i]) printf("%d ", i);
      puts("");
    } else {
      for (int i = 1; i <= n; i++) deg[i] = deg1[i];
      bool find = false;
      if (solve(k - 1, n)) {
        priority_queue<Item> q;
        for (int i = 1; i <= n; i++) {
          if (use[i] && deg[i] == k - 1) q.push(Item{i, k - 1});
        }
        while (!q.empty()) {
          Item it = q.top();
          q.pop();
          int i = it.u;
          if (deg[i] == k - 1 && check(i)) {
            puts("2");
            for (int v : e[i]) {
              if (use[v]) printf("%d ", v);
            }
            printf("%d\n", i);
            find = true;
            break;
          } else {
            for (int v : e[i]) {
              deg[i]--;
              deg[v]--;
              if (use[v] && deg[v] == k - 1) q.push(Item{v, deg[v]});
              use[i] = false;
            }
          }
        }
      }
      if (!find) {
        puts("-1");
      }
    }
    for (int i = 1; i <= n; i++) {
      e[i].clear();
    }
  }
  return 0;
}