Problem: Note that the only difference between the easy and hard version is the constraint on the number of queries. You can make hacks only if all versions of the problem are solved.

This is an interactive problem.

You are given a tree consisting of n nodes numbered with integers from 1 to n. Ayush and Ashish chose two secret distinct nodes in the tree. You need to find out both the nodes. You can make the following query: 

  * Provide a list of nodes and you will receive a node from that list whose sum of distances to both the hidden nodes is minimal (if there are multiple such nodes in the list, you will receive any one of them). You will also get the sum of distances of that node to the hidden nodes. 



Recall that a tree is a connected graph without cycles. The distance between two nodes is defined as the number of edges in the simple path between them.

More formally, let's define two hidden nodes as s and f. In one query you can provide the set of nodes \\{a_1, a_2, …, a_c\} of the tree. As a result, you will get two numbers a_i and dist(a_i, s) + dist(a_i, f). The node a_i is any node from the provided set, for which the number dist(a_i, s) + dist(a_i, f) is minimal.

You can ask no more than 11 queries.

Input

The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Please note, how the interaction process is organized.

The first line of each test case consists of a single integer n (2 ≤ n ≤ 1000) — the number of nodes in the tree.

The next n - 1 lines consist of two integers u, v (1 ≤ u, v ≤ n, u ≠ v) — the edges of the tree.

Interaction

To ask a query print a single line: 

  * In the beginning print "? c " (without quotes) where c (1 ≤ c ≤ n) denotes the number of nodes being queried, followed by c distinct integers in the range [1, n] — the indices of nodes from the list. 



For each query, you will receive two integers x, d — the node (among the queried nodes) with the minimum sum of distances to the hidden nodes and the sum of distances from that node to the hidden nodes. If the subset of nodes queried is invalid or you exceeded the number of queries then you will get x = d = -1. In this case, you should terminate the program immediately.

When you have guessed the hidden nodes, print a single line "! " (without quotes), followed by two integers in the range [1, n] — the hidden nodes. You can output the hidden nodes in any order.

After this, you should read a string. If you guess the nodes correctly, you will receive the string "Correct". In this case, you should continue solving the remaining test cases or terminate the program, if all test cases were solved. Otherwise, you will receive the string "Incorrect". In this case, you should terminate the program immediately.

Guessing the hidden nodes does not count towards the number of queries asked.

The interactor is not adaptive. The hidden nodes do not change with queries.

Do not forget to read the string "Correct" / "Incorrect" after guessing the hidden nodes.

You need to solve each test case before receiving the input for the next test case.

The limit of 11 queries applies to each test case and not to the entire input.

After printing a query do not forget to output the end of the 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 the documentation for other languages.



Hacks

To hack the solution, use the following test format:

The first line should contain a single integer t (1 ≤ t ≤ 10) — the number of test cases. The description of the test cases follows.

The first line of each test case should contain a single integer n (2 ≤ n ≤ 1000) — the number of nodes in the tree. The second line should contain two distinct integers in the range [1, n] — the hidden nodes. The next n - 1 lines should contain two integers u, v (1 ≤ u, v ≤ n, u ≠ v) — the edges of the tree.

Example

Input


1
3
1 2
1 3

1 1

2 3

3 1

3 1

Correct

Output


? 1 1

? 1 2

? 1 3

? 2 2 3

! 1 3

Note

The tree from the first test is shown below, and the hidden nodes are 1 and 3.

<image>

Solution (in JAVA):

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

import static java.lang.Math.*;
import static java.util.Arrays.*;

//

public class cf1370f2 {

    public static void main(String[] args) throws IOException {
        int t = ri();
        while (t --> 0) {
            int n = ri();
            Graph g = tree(n);
            List<Integer> qry = new ArrayList<>();
            for (int i = 0; i < n; ++i) {
                qry.add(i + 1);
            }
            int[] qans = qry(qry), h = new int[n], p = new int[n];
            int root = qans[0] - 1, d = qans[1], u = -1, v;
            p[root] = -1;
            boolean[] mark = new boolean[n];
            dfs(g, root, p, h);
            for (int l = (d + 1) / 2, r = min(maxof(h), d) + 1; l < r; ) {
                int m = l + (r - l) / 2;
                qans = qry(get(mark, h, m));
                if (qans[1] == d) {
                    u = qans[0];
                    l = m + 1;
                } else {
                    r = m;
                }
            }
            mark(u - 1, p, root, mark);
            v = qry(get(mark, h, d - h[u - 1]))[0];
            if (!ans(u, v)) {
                break;
            }
        }
        close();
    }

    static void mark(int i, int p[], int root, boolean[] mark) {
        mark[i] = true;
        if (p[i] != root) {
            mark(p[i], p, root, mark);
        }
    }

    static List<Integer> get(boolean[] mark, int[] h, int d) {
        List<Integer> ans = new ArrayList<>();
        int n = h.length;
        for (int i = 0; i < n; ++i) {
            if (!mark[i] && h[i] == d) {
                ans.add(i + 1);
            }
        }
        return ans;
    }

    static void dfs(Graph g, int i, int[] p, int[] h) {
        if (p[i] != -1) {
            h[i] = h[p[i]] + 1;
        }
        for (int j : g.get(i)) {
            if (j != p[i]) {
                p[j] = i;
                dfs(g, j, p, h);
            }
        }
    }

    static int[] qry(List<Integer> qry) throws IOException {
        pr("? ");
        pr(qry.size());
        pr(' ');
        prln(qry);
        flush();
        return ria(2);
    }

    static boolean ans(int a, int b) throws IOException {
        pr("! ");
        pr(a);
        pr(' ');
        prln(b);
        flush();
        return "Correct".equals(rline());
    }

    static Graph graph(int n) {
        Graph g = new Graph();
        for (int i = 0; i < n; ++i) {
            g.add(new ArrayList<>());
        }
        return g;
    }

    static Graph graph(int n, int m) throws IOException {
        Graph g = graph(n);
        for (int i = 0; i < m; ++i) {
            g.c(rni() - 1, ni() - 1);
        }
        return g;
    }

    static Graph tree(int n) throws IOException {
        return graph(n, n - 1);
    }

    static class Graph extends ArrayList<List<Integer>> {
        void cto(int u, int v) {
            get(u).add(v);
        }

        void c(int u, int v) {
            cto(u, v);
            cto(v, u);
        }
    }

    static BufferedReader __i = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter __o = new PrintWriter(new OutputStreamWriter(System.out));
    static StringTokenizer input;
    static Random __r = new Random();

    // references
    // IBIG = 1e9 + 7
    // IMAX ~= 2e9
    // LMAX ~= 9e18

    // constants
    static final int IBIG = 1000000007;
    static final int IMAX = 2147483647;
    static final int IMIN = -2147483648;
    static final long LMAX = 9223372036854775807L;
    static final long LMIN = -9223372036854775808L;
    // math util
    static int minof(int a, int b, int c) {return min(a, min(b, c));}
    static int minof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); int min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
    static long minof(long a, long b, long c) {return min(a, min(b, c));}
    static long minof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); long min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
    static int maxof(int a, int b, int c) {return max(a, max(b, c));}
    static int maxof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); int max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
    static long maxof(long a, long b, long c) {return max(a, max(b, c));}
    static long maxof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); long max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
    static int powi(int a, int b) {if (a == 0) return 0; int ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static long powl(long a, int b) {if (a == 0) return 0; long ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static int fli(double d) {return (int) d;}
    static int cei(double d) {return (int) ceil(d);}
    static long fll(double d) {return (long) d;}
    static long cel(double d) {return (long) ceil(d);}
    static int gcf(int a, int b) {return b == 0 ? a : gcf(b, a % b);}
    static long gcf(long a, long b) {return b == 0 ? a : gcf(b, a % b);}
    static int[] exgcd(int a, int b) {if (b == 0) return new int[] {1, 0}; int[] y = exgcd(b, a % b); return new int[] {y[1], y[0] - y[1] * (a / b)};}
    static long[] exgcd(long a, long b) {if (b == 0) return new long[] {1, 0}; long[] y = exgcd(b, a % b); return new long[] {y[1], y[0] - y[1] * (a / b)};}
    static int randInt(int min, int max) {return __r.nextInt(max - min + 1) + min;}
    static long mix(long x) {x += 0x9e3779b97f4a7c15L; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L; x = (x ^ (x >> 27)) * 0x94d049bb133111ebL; return x ^ (x >> 31);}
    // array util
    static void reverse(int[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {int swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(long[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {long swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(double[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {double swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(char[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {char swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void shuffle(int[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void shuffle(long[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void shuffle(double[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void rsort(int[] a) {shuffle(a); sort(a);}
    static void rsort(long[] a) {shuffle(a); sort(a);}
    static void rsort(double[] a) {shuffle(a); sort(a);}
    static int[] copy(int[] a) {int[] ans = new int[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static long[] copy(long[] a) {long[] ans = new long[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static double[] copy(double[] a) {double[] ans = new double[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static char[] copy(char[] a) {char[] ans = new char[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    // input
    static void r() throws IOException {input = new StringTokenizer(rline());}
    static int ri() throws IOException {return Integer.parseInt(rline());}
    static long rl() throws IOException {return Long.parseLong(rline());}
    static double rd() throws IOException {return Double.parseDouble(rline());}
    static int[] ria(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni(); return a;}
    static int[] riam1(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1; return a;}
    static long[] rla(int n) throws IOException {long[] a = new long[n]; r(); for (int i = 0; i < n; ++i) a[i] = nl(); return a;}
    static double[] rda(int n) throws IOException {double[] a = new double[n]; r(); for (int i = 0; i < n; ++i) a[i] = nd(); return a;}
    static char[] rcha() throws IOException {return rline().toCharArray();}
    static String rline() throws IOException {return __i.readLine();}
    static String n() {return input.nextToken();}
    static int rni() throws IOException {r(); return ni();}
    static int ni() {return Integer.parseInt(n());}
    static long rnl() throws IOException {r(); return nl();}
    static long nl() {return Long.parseLong(n());}
    static double rnd() throws IOException {r(); return nd();}
    static double nd() {return Double.parseDouble(n());}
    // output
    static void pr(int i) {__o.print(i);}
    static void prln(int i) {__o.println(i);}
    static void pr(long l) {__o.print(l);}
    static void prln(long l) {__o.println(l);}
    static void pr(double d) {__o.print(d);}
    static void prln(double d) {__o.println(d);}
    static void pr(char c) {__o.print(c);}
    static void prln(char c) {__o.println(c);}
    static void pr(char[] s) {__o.print(new String(s));}
    static void prln(char[] s) {__o.println(new String(s));}
    static void pr(String s) {__o.print(s);}
    static void prln(String s) {__o.println(s);}
    static void pr(Object o) {__o.print(o);}
    static void prln(Object o) {__o.println(o);}
    static void prln() {__o.println();}
    static void pryes() {prln("yes");}
    static void pry() {prln("Yes");}
    static void prY() {prln("YES");}
    static void prno() {prln("no");}
    static void prn() {prln("No");}
    static void prN() {prln("NO");}
    static boolean pryesno(boolean b) {prln(b ? "yes" : "no"); return b;};
    static boolean pryn(boolean b) {prln(b ? "Yes" : "No"); return b;}
    static boolean prYN(boolean b) {prln(b ? "YES" : "NO"); return b;}
    static void prln(int... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static void prln(long... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static void prln(double... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for (int i = 0; i < n; pr(iter.next()), pr(' '), ++i); if (n >= 0) prln(iter.next()); else prln();}
    static void h() {prln("hlfd"); flush();}
    static void flush() {__o.flush();}
    static void close() {__o.close();}
}