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 Mn = 1e5 + 5;
vector<int> G[Mn];
int du[Mn], vdel[Mn];
long long k;
inline int read() {
  int f = 1, x = 0;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch > '9' || ch < '0');
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return x * f;
}
inline void write(int x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
void Del(int n) {
  if (vdel[n]) return;
  vdel[n] = 1;
  for (int a : G[n]) {
    if (vdel[a]) continue;
    du[a]--;
    if (du[a] < k - 1) Del(a);
  }
  du[n] = 0;
  return;
}
bool pan(int n) {
  vector<int> x;
  x.push_back(n);
  for (int a : G[n]) {
    if (vdel[a]) continue;
    x.push_back(a);
  }
  for (int a : G[n]) {
    if (vdel[a]) continue;
    int t = 0;
    for (int b : x) {
      if (b == a) continue;
      if (binary_search(G[a].begin(), G[a].end(), b) == 0) return 0;
    }
  }
  return 1;
}
queue<int> q;
void Del2(int n) {
  if (vdel[n]) return;
  vdel[n] = 1;
  du[n] = 0;
  for (int it : G[n]) {
    if (vdel[it]) continue;
    du[it]--;
    if (du[it] == k - 1) q.push(it);
    if (du[it] < k - 1) Del2(it);
  }
  return;
}
inline void Slove() {
  long long n, m;
  scanf("%lld%lld%lld", &n, &m, &k);
  for (int i = 1; i <= n; i++) du[i] = vdel[i] = 0;
  int a, b;
  for (int i = 1; i <= m; i++) {
    a = read(), b = read();
    G[a].push_back(b);
    G[b].push_back(a);
    du[a]++;
    du[b]++;
  }
  if (k == 1 && k == 2) {
    puts("1 2");
    write(a);
    putchar(' ');
    write(b);
    puts("");
    for (int i = 1; i <= n; i++) G[i].clear();
    return;
  }
  if (m < (k - 1) * k / 2LL) {
    puts("-1");
    for (int i = 1; i <= n; i++) G[i].clear();
    return;
  }
  for (int i = 1; i <= n; i++) {
    sort(G[i].begin(), G[i].end());
    if (du[i] < k - 1) {
      Del(i);
    }
  }
  while (!q.empty()) q.pop();
  for (int i = 1; i <= n; i++) {
    if (du[i] == k - 1) {
      q.push(i);
    }
  }
  int t;
  while (!q.empty()) {
    t = q.front();
    q.pop();
    if (vdel[t]) continue;
    if (du[t] != k - 1) {
      Del2(t);
      continue;
    }
    if (pan(t)) {
      printf("2\n%d ", t);
      for (int it : G[t]) {
        if (vdel[it]) continue;
        write(it);
        putchar(' ');
      }
      puts("");
      for (int i = 1; i <= n; i++) G[i].clear();
      return;
    }
    Del2(t);
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (vdel[i]) continue;
    ans++;
  }
  if (ans == 0) {
    puts("-1");
    for (int i = 1; i <= n; i++) G[i].clear();
    return;
  }
  printf("1 %d\n", ans);
  for (int i = 1; i <= n; i++) {
    if (vdel[i]) continue;
    write(i);
    putchar(' ');
  }
  puts("");
  for (int i = 1; i <= n; i++) G[i].clear();
  return;
}
int main() {
  int _;
  scanf("%d", &_);
  while (_--) Slove();
  return 0;
}