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:
import java.io.*;
import java.util.*;
import java.lang.*;

public class Main {
    public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    static long MOD = (long) (1e9 + 7);
    //static int MOD = 998244353;
    static long MOD2 = MOD * MOD;
    static FastReader sc = new FastReader();
    static int pInf = Integer.MAX_VALUE;
    static int nInf = Integer.MIN_VALUE;
    static long ded = (long)(1e17)+9;
    public static void main(String[] args) throws Exception {
        int test = 1;
        test = sc.nextInt();
        for (int i = 1; i <= test; i++){
            //out.print("Case #"+i+": ");
            solve();
        }
        out.flush();
        out.close();
    }
    static void solve() throws Exception {
        int n = sc.nextInt();
        int ans = 0;
        for(int bit = 0; bit < 9; bit++){
            ArrayList<Integer> A = new ArrayList<>();
            ArrayList<Integer> B = new ArrayList<>();
            for(int i = 0; i < n; i++){
                if(((i>>bit)&1)>0){
                    A.add(i);
                }else{
                    B.add(i);
                }
            }
            if(A.size()==0||B.size()==0){
                continue;
            }
            out.print(A.size()+" "+B.size());
            for(int x: A){
                out.print(" "+(x+1));
            }
            for(int x: B){
                out.print(" "+(x+1));
            }
            out.println("");
            out.flush();
            int temp = sc.nextInt();
            ans = Math.max(ans,temp);
        }
        out.println(-1+" "+ans);
        out.flush();
    }
    static class Pair implements Comparable<Pair> {
        int x;
        long y;
        int idx;
        public Pair(int x, long y) {
            this.x = x;
            this.y = y;
        }
        @Override
        public int compareTo(Pair o) {
            if ((this.x == o.x)) {
                //return (this.y - o.y);
            }
            return (this.x - o.x);
        }
    }
    public static long mul(long a, long b) {
        return ((a % MOD) * (b % MOD)) % MOD;
    }
    public static long add(long a, long b) {
        return ((a % MOD) + (b % MOD)) % MOD;
    }
    public static long c2(long n) {
        if ((n & 1) == 0) {
            return mul(n / 2, n - 1);
        } else {
            return mul(n, (n - 1) / 2);
        }
    }
    //Shuffle Sort
    static final Random random = new Random();
    static void ruffleSort(int[] a) {
        int n = a.length;//shuffle, then sort
        for (int i = 0; i < n; i++) {
            int oi = random.nextInt(n); int temp= a[oi];
            a[oi] = a[i];
            a[i] = temp;
        }
        Arrays.sort(a);
    }
    //Brian Kernighans Algorithm
    static long countSetBits(long n) {
        if (n == 0) return 0;
        return 1 + countSetBits(n & (n - 1));
    }
    //Euclidean Algorithm
    static long gcd(long A, long B) {
        if (B == 0) return A;
        return gcd(B, A % B);
    }
    //Modular Exponentiation
    static long fastExpo(long x, long n) {
        if (n == 0) return 1;
        if ((n & 1) == 0) return fastExpo((x * x) % MOD, n / 2) % MOD;
        return ((x % MOD) * fastExpo((x * x) % MOD, (n - 1) / 2)) % MOD;
    }
    //AKS Algorithm
    static boolean isPrime(long n) {
        if (n <= 1) return false;
        if (n <= 3) return true;
        if (n % 2 == 0 || n % 3 == 0) return false;
        for (int i = 5; i <= Math.sqrt(n); i += 6)
            if (n % i == 0 || n % (i + 2) == 0) return false;
        return true;
    }
    public static long modinv(long x) {
        return modpow(x, MOD - 2);
    }
    public static long modpow(long a, long b) {
        if (b == 0) {
            return 1;
        }
        long x = modpow(a, b / 2);
        x = (x * x) % MOD;
        if (b % 2 == 1) {
            return (x * a) % MOD;
        }
        return x;
    }
    static class FastReader {
         BufferedReader br;
         StringTokenizer st;

         public FastReader() {
             br = new BufferedReader(new InputStreamReader(System.in));
         }

         String next() {
             while (st == null || !st.hasMoreElements()) {
                 try {
                     st = new StringTokenizer(br.readLine());
                 } catch (IOException e) {
                     e.printStackTrace();
                 }
             }
             return st.nextToken();
         }

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

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

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

         String nextLine() {
             String str = "";
             try {
                 str = br.readLine();
             } catch (IOException e) {
                 e.printStackTrace();
             }
             return str;
         }
     }
}