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.*;

public class C {

	public static Reader sc = new Reader();
	//public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

	public static void main(String[] args) throws IOException {
		//BufferedReader br = new BufferedReader(new FileReader("input.in"));

		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int max = 0;
			for (int i = 0; i < 9; i++) {
				HashSet<Integer> s1 = new HashSet<Integer>();
				HashSet<Integer> s2 = new HashSet<Integer>();
				for (int j = 1; j <= n; j++) {
					if ((j & (1 << i)) > 0) s1.add(j);
					else s2.add(j);
				}
				if (s1.size() > 0 && s2.size()>0) {
					System.out.print(s1.size() + " " + s2.size()  +" ");
					for (int p: s1) System.out.print(p +" ");
					for (int p: s2) System.out.print(p + " ");
					System.out.println();
					System.out.flush();
					max = Integer.max(max, sc.nextInt());
				}
				else {
					// Query s1 independently
					System.out.println(1 + " " + (n-1) + " ");
					for (int p: s1) System.out.print(p + " ");
					for (int as = 1; as <= n; as++) {
						if (!s1.contains(as)) System.out.print(as + " ");
					}
					System.out.println();
					System.out.flush();
					max = Integer.max(max, sc.nextInt());
				}
				
			}
			System.out.println(-1 + " " + max);
		}

	}

	static long ceil(long a, long b) {
		return (a + b - 1) / b;
	}

	static long powMod(long base, long exp, long mod) {
		if (base == 0 || base == 1)
			return base;
		if (exp == 0)
			return 1;
		if (exp == 1)
			return base % mod;
		long R = powMod(base, exp / 2, mod) % mod;
		R *= R;
		R %= mod;
		if ((exp & 1) == 1) {
			return base * R % mod;
		} else
			return R % mod;
	}

	static long pow(long base, long exp) {
		if (base == 0 || base == 1)
			return base;
		if (exp == 0)
			return 1;
		if (exp == 1)
			return base;
		long R = pow(base, exp / 2);
		if ((exp & 1) == 1) {
			return R * R * base;
		} else
			return R * R;
	}

	static class Reader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;

		public Reader() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public Reader(String file_name) throws IOException {
			din = new DataInputStream(new FileInputStream(file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public String readLine() throws IOException {
			byte[] buf = new byte[64]; // line length 
			int cnt = 0, c;
			while ((c = read()) != -1) {
				if (c == '\n') {
					break;
				}
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, Integer.max(cnt - 1, 0));
		}

		public int nextInt() throws IOException {
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}

		public long nextLong() throws IOException {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}

		public double nextDouble() throws IOException {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (c == '.') {
				while ((c = read()) >= '0' && c <= '9') {
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}

		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}

		private byte read() throws IOException {
			if (bufferPointer == bytesRead)
				fillBuffer();
			return buffer[bufferPointer++];
		}

		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}
	}

}