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;
void reduce(vector<long long>& taken, vector<long long>& t,
            vector<long long>& deg, vector<vector<long long> >& child,
            queue<long long>& q, long long k, queue<long long>& big) {
  while (!q.empty()) {
    long long x = q.front();
    q.pop();
    for (auto i : child[x]) {
      deg[i]--;
      if (taken[i] == 1) {
        continue;
      }
      if (deg[i] < k - 1) {
        q.push(i);
        taken[i] = 1;
      }
      if (t[i] == 1) {
        continue;
      }
      if (deg[i] == k - 1) {
        big.push(i);
        t[i] = 1;
      }
    }
  }
}
bool check(unordered_set<long long>& ny, vector<long long>& wh, long long n) {
  for (int i = 0; i < wh.size(); i++) {
    for (int j = i + 1; j < wh.size(); j++) {
      long long x = wh[i] * n + wh[j];
      if (ny.find(x) == ny.end()) {
        return false;
      }
    }
  }
  return true;
}
bool klikk(unordered_set<long long>& ny, vector<long long>& t,
           vector<long long>& taken, vector<long long>& deg,
           vector<vector<long long> >& children, queue<long long>& q,
           queue<long long>& big, long long k, vector<long long>& neigh,
           long long n) {
  while (!big.empty()) {
    long long x = big.front();
    big.pop();
    if (taken[x] == 1) {
      continue;
    }
    long long curr = 0;
    for (int i = 0; i < children[x].size(); i++) {
      if (taken[children[x][i]] == 1) {
        continue;
      }
      neigh[curr] = children[x][i];
      curr++;
    }
    if (check(ny, neigh, n)) {
      cout << 2 << endl;
      cout << x + 1 << " ";
      for (int i = 0; i < k - 1; i++) {
        cout << neigh[i] + 1 << " ";
      }
      cout << endl;
      return false;
    }
    deg[x]--;
    taken[x] = 1;
    q.push(x);
    reduce(taken, t, deg, children, q, k, big);
  }
  return true;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, m, k;
    cin >> n >> m >> k;
    vector<long long> tom;
    vector<vector<long long> > child(n, tom);
    vector<long long> deg(n, 0);
    vector<long long> taken(n, 0);
    vector<long long> t(n, 0);
    queue<long long> q;
    queue<long long> big;
    for (int i = 0; i < m; i++) {
      long long a, b;
      cin >> a >> b;
      child[a - 1].push_back(b - 1);
      child[b - 1].push_back(a - 1);
      deg[a - 1]++;
      deg[b - 1]++;
    }
    for (int i = 0; i < n; i++) {
      if (deg[i] < k - 1) {
        q.push(i);
        taken[i] = 1;
      }
      if (deg[i] == k - 1) {
        big.push(i);
        t[i] = 1;
      }
    }
    reduce(taken, t, deg, child, q, k, big);
    unordered_set<long long> ny;
    vector<vector<long long> > children(n, tom);
    for (int i = 0; i < child.size(); i++) {
      if (taken[i] == 1) {
        continue;
      }
      for (int j = 0; j < child[i].size(); j++) {
        if (taken[child[i][j]] == 0) {
          ny.insert(i * n + child[i][j]);
          ny.insert(child[i][j] * n + i);
          children[i].push_back(child[i][j]);
        }
      }
    }
    vector<long long> tomme(k - 1, 0);
    if (klikk(ny, t, taken, deg, children, q, big, k, tomme, n)) {
      long long sum = 0;
      for (int i = 0; i < n; i++) {
        sum += 1 - taken[i];
      }
      if (sum != 0) {
        cout << 1 << " " << sum << endl;
        for (int i = 0; i < n; i++) {
          if (taken[i] == 1) {
            continue;
          }
          cout << i + 1 << " ";
        }
        cout << endl;
        continue;
      }
      cout << -1 << endl;
    }
  }
}