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 JAVA):

Solution:
//package com.company;
// Always comment out package when submitting.

import java.io.*;
import java.util.*;

public class Main {
    public static class Task {
        Scanner sc;
        public int ask(int v1, List<Integer> v2) throws IOException {
            StringBuilder sb = new StringBuilder();
            sb.append(1 + " ").append(v2.size());
            sb.append(" ").append(v1 + 1);
            for (int y: v2) {
                sb.append(" ").append(y + 1);
            }
            System.out.println(sb.toString());
            System.out.flush();
            return sc.nextInt();
        }

        public void solve(Scanner sc) throws IOException {
            this.sc = sc;
            int T = sc.nextInt();
            for (int _ = 0; _ < T; _++) {
                int n = sc.nextInt();
                List<Integer> ini = new ArrayList<>();
                for (int i = 1; i < n; i++) {
                    ini.add(i);
                }
                int dist = ask(0, ini);
                while (ini.size() != 1) {
                    List<Integer> firstahlf = new ArrayList<>();
                    for (int i = 0; i < ini.size() / 2; i++) {
                        firstahlf.add(ini.get(i));
                    }
                    int size = ask(0, firstahlf);
                    if (size == dist) {
                        ini = firstahlf;
                    } else {
                        List<Integer> ses = new ArrayList<>();
                        for (int i = ini.size()/ 2; i < ini.size(); i++) {
                            ses.add(ini.get(i));
                        }
                        ini = ses;
                    }
                }
                int fur = ini.get(0);
                List<Integer> asa = new ArrayList<>();
                for (int i = 0; i < n; i++) {
                    if (i != fur) asa.add(i);
                }
                int dd = ask(fur, asa);
                System.out.println(-1 + " " + dd);
            }

        }
    }

    // template, actual code is in class Task.
    static long TIME_START, TIME_END;

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
//        Scanner sc = new Scanner(new FileInputStream("Test.in"));
//        PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));
//        PrintWriter pw = new PrintWriter(new FileOutputStream("Test.out"));

//        createTestCases();
        Runtime runtime = Runtime.getRuntime();
        long usedMemoryBefore = runtime.totalMemory() - runtime.freeMemory();
        TIME_START = System.currentTimeMillis();
        Task t = new Task();
        t.solve(sc);
        TIME_END = System.currentTimeMillis();
        long usedMemoryAfter = runtime.totalMemory() - runtime.freeMemory();
//        pw.close();
//        System.err.println("Memory increased: " + (usedMemoryAfter - usedMemoryBefore) / 1000000);
//        System.err.println("Time used: " + (TIME_END - TIME_START) + ".");
    }

    // Faster IO with BufferedReader wrapped with Scanner
    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(FileReader s) throws FileNotFoundException {
            br = new BufferedReader(s);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}