Problem: There is a weighted tree with n nodes and n-1 edges. The nodes are conveniently labeled from 1 to n. The weights are positive integers at most 100. Define the distance between two nodes to be the sum of edges on the unique path between the nodes. You would like to find the diameter of the tree. Diameter is the maximum distance between a pair of nodes.

Unfortunately, the tree isn't given to you, but you can ask some questions about it. In one question, you can specify two nonempty disjoint sets of nodes p and q, and the judge will return the maximum distance between a node in p and a node in q. In the words, maximum distance between x and y, where x ∈ p and y ∈ q. After asking not more than 9 questions, you must report the maximum distance between any pair of nodes.

Interaction

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1 000). Description of the test cases follows.

The first line of each test case contains an integer n (2 ≤ n ≤ 100) — the number of nodes in the tree.

To ask a question, print "k_1\ k_2\ a_1\ a_2\ …\ a_{k_1}\ b_1\ b_2\ …\ b_{k_2}" (k_1, k_2 ≥ 1 and k_1+k_2 ≤ n). These two sets must be nonempty and disjoint. The judge will respond with a single integer max_{p,q} dist(a_p, b_q). If you ever get a result of -1 (because you printed an invalid query), exit immediately to avoid getting other verdicts.

After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:

  * fflush(stdout) or cout.flush() in C++;
  * System.out.flush() in Java;
  * flush(output) in Pascal;
  * stdout.flush() in Python;
  * see documentation for other languages.



When you are ready to answer, print "-1\ d", where d is the maximum shortest distance over all pairs of nodes.

You can only ask at most 9 questions per test case.

Hack Format

To hack, use the following format. Note that you can only hack with one test case.

The first line should contain a single integer t (t=1).

The second line should contain a single integer n (2 ≤ n ≤ 100).

Each of the next n-1 lines should contain three integers a_i, b_i, c_i (1≤ a_i, b_i≤ n, 1 ≤ c_i ≤ 100). This denotes an undirected edge between nodes a_i and b_i with weight c_i. These edges must form a tree.

Example

Input


2
5
9
6
10
9
10
2
99


Output


1 4 1 2 3 4 5
1 4 2 3 4 5 1
1 4 3 4 5 1 2
1 4 4 5 1 2 3
1 4 5 1 2 3 4
-1 10
1 1 1 2
-1 99

Note

In the first example, the first tree looks as follows: <image>

In the first question, we have p = {1}, and q = {2, 3, 4, 5}. The maximum distance between a node in p and a node in q is 9 (the distance between nodes 1 and 5).

The second tree is a tree with two nodes with an edge with weight 99 between them.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long mod = (long long)1 << 50;
long long QPow(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans % mod * a % mod;
    a = a % mod * a % mod;
    b >>= 1;
  }
  return ans;
}
long long gcd(long long x, long long y) {
  if (x == 1 || y == 1)
    return 1;
  else {
    if (x == 0) return y;
    if (y == 0) return x;
    if (x % y == 0) return y;
    if (y % x == 0) return x;
    if (x > y)
      return gcd(x % y, y);
    else
      return gcd(x, y % x);
  }
}
string toString(long long a) {
  if (a == 0) return "0";
  string temp = "";
  if (a < 0) temp.push_back('-');
  long long tp = a;
  while (tp) {
    temp.push_back('0' + tp % 10);
    tp /= 10;
  }
  string temp2 = temp;
  for (int ii = 0; ii < temp.size(); ii++) {
    temp[ii] = temp2[temp.size() - 1 - ii];
  }
  return temp;
}
long long inf = INT_MAX;
const int maxn = 1e2 + 5;
int cnt = 0;
int head[maxn], dep[maxn], fa[maxn][22];
long long dis[maxn];
struct edge {
  int v, nx;
  long long val;
} es[maxn << 1];
void add(int u, int v, int w) {
  es[cnt].v = v;
  es[cnt].nx = head[u];
  es[cnt].val = w;
  head[u] = cnt++;
}
void dfs0(int u, int f, long long ds) {
  dis[u] = ds;
  for (int i = head[u]; i != -1; i = es[i].nx) {
    int v = es[i].v;
    if (v == f) continue;
    dfs0(v, u, ds + es[i].val);
  }
}
void dfs(int u, int f) {
  dep[u] = dep[f] + 1;
  fa[u][0] = f;
  for (int i = 1; (1 << i) <= dep[u]; i++) {
    fa[u][i] = fa[fa[u][i - 1]][i - 1];
  }
  for (int i = head[u]; i != -1; i = es[i].nx) {
    int v = es[i].v;
    if (v != f) dfs(v, u);
  }
}
int lca(int a, int b) {
  if (dep[a] > dep[b]) swap(a, b);
  for (int i = 20; i >= 0; i--)
    if (dep[a] <= dep[b] - (1 << i)) b = fa[b][i];
  if (a == b) return a;
  for (int i = 20; i >= 0; i--) {
    if (fa[a][i] == fa[b][i])
      continue;
    else
      a = fa[a][i], b = fa[b][i];
  }
  return fa[a][0];
}
set<int> ss[11][2];
void dfs(int l, int r, int cur) {
  if (l == r) return;
  int md = (l + r) >> 1;
  for (int i = l; i <= md; i++) ss[cur][0].insert(i);
  for (int i = md + 1; i <= r; i++) ss[cur][1].insert(i);
  dfs(l, md, cur + 1);
  dfs(md + 1, r, cur + 1);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ifstream in("Text.txt");
  long long k, s, t, u, v, w, q, n;
  long long d, m;
  cin >> t;
  while (t--) {
    cin >> n;
    long long ans = 0;
    dfs(1, n, 0);
    long long temp = 0;
    for (int i = 0; ss[i][0].size(); i++) {
      cout << ss[i][0].size() << " " << ss[i][1].size() << " ";
      for (auto it : ss[i][0]) cout << (it) << " ";
      for (auto it : ss[i][1]) cout << (it) << " ";
      cout.flush();
      cin >> temp;
      ans = max(ans, temp);
    }
    cout << -1 << " " << ans << endl;
    cout.flush();
    for (int i = 0; i < 11; i++) ss[i][0].clear(), ss[i][1].clear();
  }
  return 0;
}