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>
#pragma comment(linker, "/STACK:10000000000")
using namespace std;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
template <typename T>
ostream& operator<<(ostream& o, vector<T> x) {
  for (auto it : x) o << it << ' ';
  return o;
}
template <typename T>
istream& operator>>(istream& i, vector<T>& x) {
  for (auto& it : x) i >> it;
  return i;
}
template <typename T1, typename T2>
void chkmin(T1& a, T2 b) {
  a > b ? a = b : 0;
}
template <typename T1, typename T2>
void chkmax(T1& a, T2 b) {
  a < b ? a = b : 0;
}
const long long INF = 1000000007;
const long long INF2 = 100000000000000007;
const long long mINF = 1e16;
const long long MOD1 = 998244353;
const long long MOD = 1000000007;
const long long SIZE = 200010;
const long long base = 53;
const long double eps = 1e-6;
const long double pi = acosl(-1.0);
long long add(long long a, long long b) { return ((a + b) % MOD + MOD) % MOD; }
long long mult(long long a, long long b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
long long binpow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = mult(res, a);
    a = mult(a, a);
    n >>= 1;
  }
  return res;
}
long long binpow1(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res *= a;
    }
    a *= a;
    n >>= 1;
  }
  return res;
}
long long inv(long long a) { return binpow(a, MOD - 2); }
long long del(long long a, long long b) { return mult(a, inv(b)); }
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
template <typename T>
T sqr(T x) {
  return x * x;
}
long long q, n, m, k, v, v1, deg[SIZE], deg1[SIZE];
vector<long long> g[SIZE];
bool ok;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> q;
  while (q--) {
    cin >> n >> m >> k;
    for (long long i = 0; i < n; ++i) {
      g[i].clear();
      deg[i] = 0;
    }
    for (long long i = 0; i < m; ++i) {
      cin >> v >> v1;
      v--;
      v1--;
      deg[v]++;
      deg[v1]++;
      g[v].push_back(v1);
      g[v1].push_back(v);
    }
    set<pair<long long, long long> > verts;
    for (long long i = 0; i < n; ++i)
      verts.insert(make_pair(deg[i], i)), sort(g[i].begin(), g[i].end());
    while (!verts.empty() && verts.begin()->first < k - 1) {
      v = verts.begin()->second;
      for (auto& i : g[v]) {
        if (deg[i] > 0) {
          verts.erase(make_pair(deg[i], i));
          deg[i]--;
          if (deg[i] > 0) verts.insert(make_pair(deg[i], i));
        }
      }
      verts.erase(make_pair(deg[v], v));
      deg[v] = 0;
    }
    ok = 1;
    if (verts.size() == k) {
      for (auto& i : verts) {
        ok &= (i.first == k - 1);
      }
      if (ok) {
        cout << 2 << endl;
        for (auto& i : verts) {
          cout << i.second + 1 << ' ';
        }
        cout << endl;
      }
    } else if (verts.size() > k) {
      set<pair<long long, long long> > verts1(verts);
      for (long long i = 0; i < n; ++i) deg1[i] = deg[i];
      while (!verts1.empty()) {
        v = verts1.begin()->second;
        vector<long long> adj_to_v;
        for (long long& i : g[v]) {
          if (deg1[i] > 0) adj_to_v.push_back(i);
        }
        if (deg1[v] == k - 1) {
          ok = 1;
          for (long long i = 0; i < k - 1; ++i) {
            for (long long j = i + 1; j < k - 1; ++j) {
              auto it = lower_bound(g[adj_to_v[i]].begin(),
                                    g[adj_to_v[i]].end(), adj_to_v[j]);
              if (it == g[adj_to_v[i]].end() || *it != adj_to_v[j]) {
                ok = 0;
                break;
              }
            }
            if (!ok) break;
          }
          if (ok) {
            cout << 2 << endl;
            cout << v + 1 << ' ';
            for (long long& i : adj_to_v) cout << i + 1 << ' ';
            cout << endl;
            break;
          }
        }
        ok = 0;
        for (long long& i : adj_to_v) {
          verts1.erase(make_pair(deg1[i], i));
          deg1[i]--;
          if (deg1[i] > 0) verts1.insert(make_pair(deg1[i], i));
        }
        verts1.erase(make_pair(deg1[v], v));
        deg1[v] = 0;
      }
    } else
      ok = 0;
    if (ok) continue;
    while (!verts.empty() && verts.begin()->first < k) {
      v = verts.begin()->second;
      for (auto& i : g[v]) {
        if (deg[i] > 0) {
          verts.erase(make_pair(deg[i], i));
          deg[i]--;
          if (deg[i] > 0) verts.insert(make_pair(deg[i], i));
        }
      }
      verts.erase(make_pair(deg[v], v));
      deg[v] = 0;
    }
    if (!verts.empty()) {
      cout << 1 << endl;
      cout << verts.size() << endl;
      for (auto& it : verts) cout << it.second + 1 << ' ';
      cout << endl;
      continue;
    } else
      cout << "-1\n";
  }
  return 0;
}