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;
template <typename _Tp>
inline void red(_Tp &x) {
  x = 0;
  bool fg = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') fg ^= 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + (_Tp)(ch ^ 48), ch = getchar();
  if (fg) x = -x;
}
template <typename _Tp>
bool umax(_Tp &x, _Tp y) {
  return (x < y) ? (x = y, true) : (false);
}
template <typename _Tp>
bool umin(_Tp &x, _Tp y) {
  return (x > y) ? (x = y, true) : (false);
}
const int N = 200010;
int n, m, K, T, deg[N];
int vis[N];
vector<int> e[N];
int main() {
  red(T);
  while (T--) {
    red(n);
    red(m);
    red(K);
    for (int i = 1; i <= n; ++i) e[i].clear(), deg[i] = 0, vis[i] = 0;
    for (int i = 1; i <= m; ++i) {
      int u, v;
      red(u);
      red(v);
      e[u].push_back(v);
      e[v].push_back(u);
      deg[u]++;
      deg[v]++;
    }
    for (int i = 1; i <= n; ++i) sort(e[i].begin(), e[i].end());
    int cnt = 0;
    queue<int> q;
    for (int i = 1; i <= n; ++i)
      if (deg[i] < K) q.push(i), vis[i] = 1;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      cnt++;
      for (auto it = e[u].begin(); it != e[u].end(); ++it) {
        --deg[*it];
        if (!vis[*it] && deg[*it] < K) q.push(*it), vis[*it] = 1;
      }
    }
    if (cnt != n) {
      printf("1 %d\n", n - cnt);
      for (int i = 1; i <= n; ++i)
        if (!vis[i]) printf("%d ", i);
      puts("");
      continue;
    }
    if (1ll * K * (K - 1) / 2 > m) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; ++i) {
      deg[i] = e[i].size();
      vis[i] = 0;
      if (deg[i] <= K - 1) {
        vis[i] = 1;
        q.push(i);
      }
    }
    bool ok = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      bool fg = 1;
      vis[u] = 2;
      if (deg[u] < K - 1) fg = 0;
      vector<int> tmp;
      if (fg) {
        for (int i = 0; i < e[u].size(); ++i)
          if (vis[e[u][i]] != 2) tmp.push_back(e[u][i]);
        for (int i = 0; i < tmp.size(); ++i) {
          for (int j = i + 1; j < tmp.size(); ++j) {
            if (lower_bound(e[tmp[i]].begin(), e[tmp[i]].end(), tmp[j]) ==
                    e[tmp[i]].end() ||
                *lower_bound(e[tmp[i]].begin(), e[tmp[i]].end(), tmp[j]) !=
                    tmp[j]) {
              fg = 0;
              break;
            }
          }
        }
      }
      if (fg) {
        printf("2\n%d", u);
        for (int i = 0; i < tmp.size(); ++i) printf(" %d", tmp[i]);
        puts("");
        ok = 1;
        break;
      } else {
        for (int i = 0; i < e[u].size(); ++i) {
          deg[e[u][i]]--;
          if (vis[e[u][i]] == 0 && deg[e[u][i]] == K - 1)
            q.push(e[u][i]), vis[e[u][i]] = 1;
        }
      }
    }
    if (!ok) puts("-1");
  }
  return 0;
}