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 double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int t, n, m, k, tot;
vector<int> g[maxn];
int col[maxn], del[maxn], deg[maxn], ok[maxn];
queue<int> q;
vector<int> ans, num[maxn];
vector<int> cq[maxn];
int main(void) {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i++) {
      g[i].clear();
      del[i] = 0;
      deg[i] = 0;
    }
    int u, v;
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &u, &v);
      g[u].push_back(v);
      g[v].push_back(u);
      deg[u]++;
      deg[v]++;
    }
    while (!q.empty()) q.pop();
    for (int i = 1; i <= n; i++) {
      if (deg[i] < k) {
        q.push(i);
        del[i] = 1;
      }
    }
    int cnt = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      ++cnt;
      for (auto v : g[u]) {
        if (del[v]) continue;
        deg[v]--;
        if (deg[v] < k) {
          q.push(v);
          del[v] = 1;
        }
      }
    }
    if (cnt < n) {
      ans.clear();
      for (int i = 1; i <= n; i++) {
        if (!del[i]) {
          ans.push_back(i);
        }
      }
      printf("1 %d\n", (int)(ans).size());
      for (int i = 0; i < (int)(ans).size(); i++) {
        printf("%d ", ans[i]);
      }
      puts("");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      del[i] = 0;
      deg[i] = (int)(g[i]).size();
      num[i].clear();
    }
    for (int i = 1; i <= n; i++) {
      if (deg[i] < k - 1) {
        q.push(i);
        del[i] = 1;
      }
    }
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (auto v : g[u]) {
        if (del[v]) continue;
        deg[v]--;
        if (deg[v] < k - 1) {
          del[v] = 1;
          q.push(v);
        }
      }
    }
    tot = 0;
    for (int i = 1; i <= n; i++) {
      if (deg[i] == k - 1) {
        q.push(i);
        del[i] = 2;
      }
    }
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      del[u] = 1;
      if (deg[u] == k - 1) {
        ++tot;
        cq[tot].clear();
        cq[tot].push_back(u);
        num[u].push_back(tot);
        for (auto v : g[u]) {
          if (del[v] == 1) continue;
          cq[tot].push_back(v);
          num[v].push_back(tot);
        }
      }
      for (auto v : g[u]) {
        deg[v]--;
        if (del[v]) continue;
        if (deg[v] < k) {
          q.push(v);
          del[v] = 2;
        }
      }
    }
    for (int i = 1; i <= tot; i++) ok[i] = 1;
    for (int i = 1; i <= n; i++) {
      col[i] = 1;
      for (auto v : g[i]) {
        col[v] = 1;
      }
      for (auto v : num[i]) {
        int f = 1;
        for (auto w : cq[v]) {
          if (col[w] == 0) {
            f = 0;
            break;
          }
        }
        if (!f) ok[v] = 0;
      }
      col[i] = 0;
      for (auto v : g[i]) {
        col[v] = 0;
      }
    }
    int f = 0;
    for (int i = 1; i <= tot; i++) {
      if (ok[i]) {
        puts("2");
        for (auto v : cq[i]) {
          printf("%d ", v);
        }
        puts("");
        f = 1;
        break;
      }
    }
    if (f) continue;
    puts("-1");
  }
  return 0;
}