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;
const int N = 100010;
vector<int> v[N];
unordered_set<int> edges[N];
bool instack[N];
int deg[N];
void solve() {
  int n, m, r;
  scanf("%d%d%d", &n, &m, &r);
  for (int i = 1; i <= n; i++) {
    v[i].clear();
    deg[i] = 0;
    instack[i] = true;
    edges[i].clear();
  }
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    deg[x]++, deg[y]++;
    v[x].push_back(y);
    v[y].push_back(x);
    edges[x].insert(y);
    edges[y].insert(x);
  }
  set<pair<int, int> > st;
  for (int i = 1; i <= n; i++) st.insert(make_pair(deg[i], i));
  while (!st.empty() && st.begin()->first < r) {
    int x = st.begin()->second;
    st.erase(*st.begin());
    instack[x] = false;
    for (int i = 0; i < v[x].size(); i++) {
      int y = v[x][i];
      if (!instack[y]) continue;
      st.erase(make_pair(deg[y], y));
      deg[y]--;
      st.insert(make_pair(deg[y], y));
    }
  }
  if (st.size()) {
    printf("1 %d\n", int(st.size()));
    for (set<pair<int, int> >::iterator it = st.begin(); it != st.end(); it++)
      printf("%d ", it->second);
    printf("\n");
    return;
  }
  if (1LL * r * (r - 1) / 2 > m) {
    printf("-1\n");
    return;
  }
  st.clear();
  for (int i = 1; i <= n; i++) {
    deg[i] = v[i].size();
    instack[i] = true;
    st.insert(make_pair(deg[i], i));
  }
  while (!st.empty() && st.begin()->first < r - 1) {
    int x = st.begin()->second;
    instack[x] = false;
    st.erase(*st.begin());
    for (int i = 0; i < v[x].size(); i++) {
      int y = v[x][i];
      if (!instack[y]) continue;
      st.erase(make_pair(deg[y], y));
      deg[y]--;
      st.insert(make_pair(deg[y], y));
    }
  }
  if (st.size() < r) {
    printf("-1\n");
    return;
  }
  while (st.size() >= r) {
    int x = st.begin()->second;
    int y = st.begin()->first;
    instack[x] = false;
    vector<int> tmp;
    for (int i = 0; i < v[x].size() && y == r - 1; i++)
      if (instack[v[x][i]]) {
        tmp.push_back(v[x][i]);
      }
    tmp.push_back(x);
    bool flag = true;
    for (int i = 0; i < tmp.size() && flag && y == r - 1; i++) {
      int t = tmp[i];
      for (int j = i + 1; j < tmp.size() && flag; j++) {
        int p = tmp[j];
        if (!edges[t].count(p)) {
          flag = false;
          break;
        }
      }
    }
    if (flag && y == r - 1) {
      printf("2\n");
      for (int i = 0; i < tmp.size(); i++) printf("%d ", tmp[i]);
      printf("\n");
      return;
    }
    st.erase(*st.begin());
    for (int i = 0; i < v[x].size(); i++) {
      y = v[x][i];
      if (!instack[y]) continue;
      st.erase(make_pair(deg[y], y));
      deg[y]--;
      st.insert(make_pair(deg[y], y));
    }
  }
  printf("-1\n");
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}