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;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
const long long MAXN = 1e5 + 10;
const long long INF = 0x3f3f3f3f;
const long long MOD = 1e9 + 7;
const long double eps = 1e-9;
const long double PI = acos(-1.0);
long long n, m, k;
vector<long long> E[MAXN], temp;
long long vis[MAXN], inEdge[MAXN];
struct Node {
  long long deg, idx;
  bool operator<(const Node &other) const {
    if (deg == other.deg)
      return idx < other.idx;
    else
      return deg < other.deg;
  }
};
set<Node> S;
void solve() {
  while (!S.empty()) {
    Node node = *S.begin();
    vis[node.idx] = 1;
    if (node.deg >= k) {
      cout << 1 << " " << S.size() << endl;
      for (auto it = S.begin(); it != S.end(); it++) {
        if (it != S.begin()) cout << " ";
        cout << (*it).idx;
      }
      cout << endl;
      return;
    } else {
      if (node.deg == k - 1) {
        temp.clear();
        long long u = node.idx;
        temp.push_back(u);
        for (auto v : E[u]) {
          if (vis[v]) continue;
          temp.push_back(v);
        }
        long long res = 1;
        for (auto v : temp) {
          for (auto t : temp) {
            if (v == t) continue;
            auto pos = lower_bound(E[v].begin(), E[v].end(), t);
            if (pos == E[v].end() || *pos != t) {
              res = 0;
              break;
            }
          }
          if (!res) break;
        }
        if (res) {
          cout << 2 << endl;
          for (long long i = 0; i < temp.size(); i++) {
            if (i) cout << " ";
            cout << temp[i];
          }
          cout << endl;
          return;
        }
      }
      S.erase(S.begin());
      for (auto v : E[node.idx]) {
        if (vis[v]) continue;
        S.erase({inEdge[v], v});
        S.insert({--inEdge[v], v});
      }
    }
  }
  cout << -1 << endl;
}
signed main() {
  long long T = read();
  while (T--) {
    S.clear();
    n = read(), m = read(), k = read();
    for (long long i = 0; i <= n; i++) {
      vis[i] = inEdge[i] = 0;
      E[i].clear();
    }
    for (long long i = 0; i < m; i++) {
      long long u = read(), v = read();
      E[u].push_back(v);
      E[v].push_back(u);
      inEdge[u]++, inEdge[v]++;
    }
    if (k * (k - 1) > 2 * m) {
      cout << -1 << endl;
      continue;
    }
    for (long long i = 1; i <= n; i++) {
      S.insert({inEdge[i], i});
      sort(E[i].begin(), E[i].end());
    }
    solve();
  }
  return 0;
}