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;
int N, M, K;
vector<int> adj[101010];
bool vst[101010];
int deg[101010];
queue<int> q;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
unordered_map<long long, int, custom_hash> mp;
bool isEdge(int a, int b) {
  if (a > b) swap(a, b);
  long long x = (long long)a * (N + 1) + b;
  return mp[x];
}
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    mp.clear();
    scanf("%d%d%d", &N, &M, &K);
    for (int i = 1; i <= N; i++) {
      adj[i].clear();
      deg[i] = 0;
      vst[i] = false;
    }
    while (q.size()) q.pop();
    for (int i = 1; i <= M; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      adj[a].push_back(b);
      adj[b].push_back(a);
      deg[a]++;
      deg[b]++;
      if (a > b) swap(a, b);
      mp[(long long)a * (N + 1) + b]++;
    }
    for (int i = 1; i <= N; i++) {
      if (deg[i] < K) {
        vst[i] = true;
        q.push(i);
      }
    }
    while (q.size()) {
      int x = q.front();
      q.pop();
      for (int i : adj[x]) {
        if (vst[i]) continue;
        deg[i]--;
        if (deg[i] < K) {
          vst[i] = true;
          q.push(i);
        }
      }
    }
    int cnt = 0;
    for (int i = 1; i <= N; i++)
      if (!vst[i]) cnt++;
    if (cnt) {
      printf("1 %d\n", cnt);
      for (int i = 1; i <= N; i++)
        if (!vst[i]) printf("%d ", i);
      puts("");
      continue;
    }
    if ((long long)K * (K - 1) / 2 > M) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= N; i++) {
      deg[i] = adj[i].size();
      vst[i] = false;
    }
    for (int i = 1; i <= N; i++) {
      if (deg[i] < K - 1) {
        vst[i] = true;
        q.push(i);
      }
    }
    while (q.size()) {
      int x = q.front();
      q.pop();
      for (int i : adj[x]) {
        if (vst[i]) continue;
        deg[i]--;
        if (deg[i] < K - 1) {
          vst[i] = true;
          q.push(i);
        }
      }
    }
    for (int i = 1; i <= N; i++) {
      if (deg[i] == K - 1) {
        vst[i] = true;
        q.push(i);
      }
    }
    vector<int> v;
    while (q.size()) {
      int x = q.front();
      q.pop();
      if (deg[x] < K - 1) {
        for (int i : adj[x]) {
          if (deg[i] < K - 1) continue;
          deg[i]--;
          if (!vst[i] && deg[i] <= K - 1) {
            vst[i] = true;
            q.push(i);
          }
        }
        continue;
      }
      for (int i : adj[x]) {
        if (deg[i] < K - 1) continue;
        v.push_back(i);
      }
      v.push_back(x);
      if (v.size() < K) {
        v.clear();
        deg[x] = 0;
        for (int i : adj[x]) {
          if (deg[i] < K - 1) continue;
          deg[i]--;
          if (!vst[i] && deg[i] <= K - 1) {
            vst[i] = true;
            q.push(i);
          }
        }
        continue;
      }
      bool ok = true;
      for (int i = 0; i < v.size(); i++) {
        if (!ok) break;
        for (int j = 0; j < i; j++) {
          if (!isEdge(v[i], v[j])) {
            ok = false;
            break;
          }
        }
      }
      if (ok) break;
      v.clear();
      deg[x] = 0;
      for (int i : adj[x]) {
        if (deg[i] < K - 1) continue;
        deg[i]--;
        if (!vst[i] && deg[i] <= K - 1) {
          vst[i] = true;
          q.push(i);
        }
      }
    }
    if (v.size()) {
      puts("2");
      for (int i : v) printf("%d ", i);
      puts("");
    } else
      puts("-1");
  }
  return 0;
}