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>
namespace FastIO {
bool IOerror = 0;
inline char nc() {
  static char buf[100000], *p1 = buf + 100000, *pend = buf + 100000;
  if (p1 == pend) {
    p1 = buf;
    pend = buf + fread(buf, 1, 100000, stdin);
    if (pend == p1) {
      IOerror = 1;
      return -1;
    }
  }
  return *p1++;
}
inline bool blank(char ch) {
  return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
template <class T>
inline bool read(T& x) {
  bool sign = 0;
  char ch = nc();
  x = 0;
  for (; blank(ch); ch = nc())
    ;
  if (IOerror) return false;
  if (ch == '-') sign = 1, ch = nc();
  for (; ch >= '0' && ch <= '9'; ch = nc()) x = x * 10 + ch - '0';
  if (sign) x = -x;
  return true;
}
template <class T, class... U>
bool read(T& h, U&... t) {
  return read(h) && read(t...);
}
};  // namespace FastIO
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;
int deg[N], vis[N];
vector<int> e[N];
int main() {
  int T;
  read(T);
  while (T--) {
    int n, m, k;
    read(n, m, k);
    for (int i = 1; i <= n; i++) e[i].clear(), vis[i] = 0;
    for (int i = 1; i <= m; i++) {
      int u, v;
      read(u, v);
      e[u].push_back(v);
      e[v].push_back(u);
    }
    queue<int> q;
    vector<int> clique;
    for (int i = 1; i <= n; i++) {
      deg[i] = (e[i]).size();
      sort((e[i]).begin(), (e[i]).end());
      if (deg[i] < k) {
        vis[i] = 1;
        q.push(i);
      }
    }
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      vis[u] = 2;
      if (1LL * k * (k - 1) / 2 <= 1LL * m && deg[u] == k - 1 &&
          clique.empty()) {
        int ok = 1;
        clique.push_back(u);
        for (auto v : e[u])
          if (vis[v] != 2) clique.push_back(v);
        for (auto x : clique) {
          if (!ok) break;
          for (auto y : clique) {
            if (x != y && !binary_search((e[x]).begin(), (e[x]).end(), y)) {
              ok = 0;
              break;
            }
          }
        }
        if (!ok) clique.clear();
      }
      for (auto v : e[u]) {
        --deg[v];
        if (deg[v] < k && !vis[v]) {
          q.push(v);
          vis[v] = 1;
        }
      }
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++)
      if (!vis[i]) cnt++;
    if (cnt) {
      printf("1 %d\n", cnt);
      for (int i = 1; i <= n; i++)
        if (!vis[i]) printf("%d ", i);
      puts("");
    } else if (!clique.empty()) {
      puts("2");
      for (int i = 0; i < k; i++) printf("%d ", clique[i]);
      puts("");
    } else {
      puts("-1");
    }
  }
  return 0;
}