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;
clock_t timeStart, timeFinish;
void timeBegin() { timeStart = clock(); }
void timeEnd() { timeFinish = clock(); }
void timeDuration() {
  double time_taken = double(timeFinish - timeStart) / double(CLOCKS_PER_SEC);
  cout << "Time taken by program is : " << fixed << time_taken
       << setprecision(5);
  cout << " sec " << endl;
}
class InputReader {
 public:
  InputReader() {
    input_file = stdin;
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  InputReader(const char *file_name) {
    input_file = fopen(file_name, "r");
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  inline InputReader &operator>>(int &n) {
    while ((buffer[cursor] < '0' || buffer[cursor] > '9') &&
           buffer[cursor] != '-') {
      advance();
    }
    int sign = 1;
    if (buffer[cursor] == '-') {
      sign = -1;
      advance();
    }
    n = 0;
    while ('0' <= buffer[cursor] && buffer[cursor] <= '9') {
      n = n * 10 + buffer[cursor] - '0';
      advance();
    }
    n *= sign;
    return *this;
  }
  inline InputReader &operator>>(long long &n) {
    while ((buffer[cursor] < '0' || buffer[cursor] > '9') &&
           buffer[cursor] != '-') {
      advance();
    }
    int sign = 1;
    if (buffer[cursor] == '-') {
      sign = -1;
      advance();
    }
    n = 0;
    while ('0' <= buffer[cursor] && buffer[cursor] <= '9') {
      n = n * 10 + buffer[cursor] - '0';
      advance();
    }
    n *= sign;
    return *this;
  }

 private:
  FILE *input_file;
  static const int SIZE = 1 << 17;
  int cursor;
  char buffer[SIZE];
  inline void advance() {
    ++cursor;
    if (cursor == SIZE) {
      cursor = 0;
      fread(buffer, SIZE, 1, input_file);
    }
  }
};
const int MAXN = 100000;
const int MOD = 666013;
const int BASE = 1000000007;
vector<pair<int, int> > table[MOD];
vector<int> g[1 + MAXN], sol;
int degree[1 + MAXN];
bool out[1 + MAXN], found;
int code(int a, int b) { return (1LL * a * BASE + b) % MOD; }
void erase(int a, int b) {
  if (a > b) {
    swap(a, b);
  }
  int h = code(a, b);
  table[h].clear();
}
void insert(int a, int b) {
  if (a > b) {
    swap(a, b);
  }
  int h = code(a, b);
  table[h].push_back(make_pair(a, b));
}
bool find(int a, int b) {
  if (a > b) {
    swap(a, b);
  }
  int h = code(a, b);
  for (auto it : table[h]) {
    if (it.first == a && it.second == b) {
      return true;
    }
  }
  return false;
}
void check(int node) {
  vector<int> nodes;
  for (auto it : g[node]) {
    if (!out[it]) {
      nodes.push_back(it);
    }
  }
  for (int i = 0; i < nodes.size(); i++) {
    for (int j = i + 1; j < nodes.size(); j++) {
      if (!find(nodes[i], nodes[j])) {
        return;
      }
    }
  }
  found = true;
  sol = nodes;
  sol.push_back(node);
}
int main() {
  int tests;
  cin >> tests;
  for (int test = 1; test <= tests; test++) {
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 1; i <= m; i++) {
      int a, b;
      cin >> a >> b;
      g[a].push_back(b);
      g[b].push_back(a);
      insert(a, b);
      degree[a]++;
      degree[b]++;
    }
    if (1LL * (k - 1) * (k - 2) / 2 > m) {
      cout << "-1\n";
    } else {
      found = false;
      sol.clear();
      queue<int> q;
      for (int i = 1; i <= n; i++) {
        if (degree[i] == k - 1) {
          check(i);
          if (found) {
            break;
          }
        }
        if (degree[i] < k) {
          q.push(i);
        }
      }
      while (!found && !q.empty()) {
        int node = q.front();
        q.pop();
        out[node] = true;
        for (auto it : g[node]) {
          degree[it]--;
          if (degree[it] == k - 1) {
            check(it);
            q.push(it);
          }
        }
      }
      if (found) {
        cout << "2\n";
        for (auto it : sol) {
          cout << it << " ";
        }
        cout << "\n";
      } else {
        for (int i = 1; i <= n; i++) {
          if (!out[i]) {
            sol.push_back(i);
          }
        }
        if (!sol.empty()) {
          found = true;
        }
        if (!found) {
          cout << "-1\n";
        } else {
          cout << "1 " << sol.size() << "\n";
          for (auto it : sol) {
            cout << it << " ";
          }
          cout << "\n";
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      for (auto it : g[i]) {
        erase(i, it);
      }
      g[i].clear();
      out[i] = false;
      degree[i] = 0;
    }
  }
  return 0;
}