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 cf1146;

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.InputMismatchException;

public class CFC {
    private static final String INPUT = "2\n" +
            "5\n" +
            "9\n" +
            "6\n" +
            "10\n" +
            "9\n" +
            "10\n" +
            "2\n" +
            "99";
    private static final int MOD = 1_000_000_007;

    private PrintWriter out;
    private FastScanner sc;

    public static void main(String[] args) {
        new CFC().run();
    }

    public void run() {
        sc = new FastScanner(oj ? System.in : new ByteArrayInputStream(INPUT.getBytes()));
        out = new PrintWriter(System.out);

        long s = System.currentTimeMillis();
        solve();
        out.flush();
        tr(System.currentTimeMillis() - s + "ms");
    }

    public void solve() {
        int t = sc.nextInt();

        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();

            int max = q(1, 2, n);
            int low = 2;
            int hig = n;

            while (low != hig) {
                int med = low + (hig - low) / 2;
                int cur = q(1, low, med);
                if (cur < max) {
                    low = med + 1;
                } else {
                    hig = med;
                }
            }
            out.printf("-1 %d%n", q(low, n));
            out.flush();
        }
    }

    private int q(int one, int low, int hig) {
        out.print("1 ");
        out.print((hig - low + 1) + " ");
        out.print(one + " ");
        for (int i = low; i <= hig; i++) {
            out.print(i + " ");
        }
        out.println();
        out.flush();

        return sc.nextInt();
    }

    private int q(int one, int size) {
        out.print("1 ");
        out.print((size - 1) + " ");
        out.print(one + " ");
        for (int i = 1; i <= size; i++) {
            if (i != one) out.print(i + " ");
        }
        out.println();
        out.flush();

        return sc.nextInt();
    }

    //********************************************************************************************
    //********************************************************************************************
    //********************************************************************************************

    /**
     * If searched element doesn't exist, returns index of first element which is bigger than searched value.<br>
     * If searched element is bigger than any array element function returns first index after last element.<br>
     * If searched element is lower than any array element function returns index of first element.<br>
     * If there are many values equals searched value function returns first occurrence.<br>
     */
    private static int lowerBound(long[] arr, long key) {
        int lo = 0;
        int hi = arr.length - 1;
        while (lo < hi) {
            int mid = (lo + hi) / 2;
            if (key <= arr[mid]) {
                hi = mid - 1;
            } else {
                lo = mid + 1;
            }
        }
        return arr[lo] < key ? lo + 1 : lo;
    }

    /**
     * Returns index of first element which is grater than searched value.
     * If searched element is bigger than any array element, returns first index after last element.
     */
    private static int upperBound(long[] arr, long key) {
        int lo = 0;
        int hi = arr.length - 1;
        while (lo < hi) {
            int mid = (lo + hi) / 2;
            if (key >= arr[mid]) {
                lo = mid + 1;
            } else {
                hi = mid;
            }
        }
        return arr[lo] <= key ? lo + 1 : lo;
    }

    private static int ceil(double d) {
        int ret = (int) d;
        return ret == d ? ret : ret + 1;
    }

    private static int round(double d) {
        return (int) (d + 0.5);
    }

    private static int gcd(int a, int b) {
        BigInteger b1 = BigInteger.valueOf(a);
        BigInteger b2 = BigInteger.valueOf(b);
        BigInteger gcd = b1.gcd(b2);
        return gcd.intValue();
    }

    private static long gcd(long a, long b) {
        BigInteger b1 = BigInteger.valueOf(a);
        BigInteger b2 = BigInteger.valueOf(b);
        BigInteger gcd = b1.gcd(b2);
        return gcd.longValue();
    }

    private int[] readIntArray(int n) {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = sc.nextInt();
        }
        return res;
    }

    private long[] readLongArray(int n) {
        long[] res = new long[n];
        for (int i = 0; i < n; i++) {
            res[i] = sc.nextLong();
        }
        return res;
    }

    @SuppressWarnings("unused")
    static class FastScanner {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;

        FastScanner(InputStream stream) {
            this.stream = stream;
        }

        int read() {
            if (numChars == -1)
                throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0) return -1;
            }
            return buf[curChar++];
        }

        boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        boolean isEndline(int c) {
            return c == '\n' || c == '\r' || c == -1;
        }

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

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

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

        public String next() {
            int c = read();
            while (isSpaceChar(c)) c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }

        public String nextLine() {
            int c = read();
            while (isEndline(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndline(c));
            return res.toString();
        }

    }

    private boolean oj = System.getProperty("ONLINE_JUDGE") != null;

    private void tr(Object... o) {
        if (!oj) System.out.println(Arrays.deepToString(o));
    }
}