Problem: This is an interactive problem.

After getting AC after 13 Time Limit Exceeded verdicts on a geometry problem, Kuroni went to an Italian restaurant to celebrate this holy achievement. Unfortunately, the excess sauce disoriented him, and he's now lost!

The United States of America can be modeled as a tree (why though) with n vertices. The tree is rooted at vertex r, wherein lies Kuroni's hotel.

Kuroni has a phone app designed to help him in such emergency cases. To use the app, he has to input two vertices u and v, and it'll return a vertex w, which is the lowest common ancestor of those two vertices.

However, since the phone's battery has been almost drained out from live-streaming Kuroni's celebration party, he could only use the app at most ⌊ n/2 ⌋ times. After that, the phone would die and there will be nothing left to help our dear friend! :(

As the night is cold and dark, Kuroni needs to get back, so that he can reunite with his comfy bed and pillow(s). Can you help him figure out his hotel's location?

Interaction

The interaction starts with reading a single integer n (2 ≤ n ≤ 1000), the number of vertices of the tree.

Then you will read n-1 lines, the i-th of them has two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), denoting there is an edge connecting vertices x_i and y_i. It is guaranteed that the edges will form a tree.

Then you can make queries of type "? u v" (1 ≤ u, v ≤ n) to find the lowest common ancestor of vertex u and v.

After the query, read the result w as an integer.

In case your query is invalid or you asked more than ⌊ n/2 ⌋ queries, the program will print -1 and will finish interaction. You will receive a Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.

When you find out the vertex r, print "! r" and quit after that. This query does not count towards the ⌊ n/2 ⌋ limit.

Note that the tree is fixed beforehand and will not change during the queries, i.e. the interactor is not adaptive.

After printing any query do not forget to print end of line and flush the output. Otherwise, you might 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, use the following format:

The first line should contain two integers n and r (2 ≤ n ≤ 1000, 1 ≤ r ≤ n), denoting the number of vertices and the vertex with Kuroni's hotel.

The i-th of the next n-1 lines should contain two integers x_i and y_i (1 ≤ x_i, y_i ≤ n) — denoting there is an edge connecting vertex x_i and y_i.

The edges presented should form a tree.

Example

Input


6
1 4
4 2
5 3
6 3
2 3

3

4

4



Output








? 5 6

? 3 1

? 1 2

! 4

Note

Note that the example interaction contains extra empty lines so that it's easier to read. The real interaction doesn't contain any empty lines and you shouldn't print any extra empty lines as well.

The image below demonstrates the tree in the sample test:

<image>

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.Arrays;
import java.util.StringJoiner;
import java.util.StringTokenizer;

public class MainD {

    static int N;
    static Edge[] E;

    public static void main(String[] args) {
        FastScanner sc = new FastScanner(System.in);
        N = sc.nextInt();
        E = new Edge[N-1];
        for (int i = 0; i < N - 1; i++) {
            E[i] = new Edge(sc.nextInt()-1, sc.nextInt()-1);
        }

        Query q = new Query() {
            @Override
            public int excuse(int u, int v) {
                System.out.println("? " + (u+1) + " " + (v+1));
                System.out.flush();
                int w = sc.nextInt();
                if( w == -1 ) throw new RuntimeException("wtf");
                return w-1;
            }

            @Override
            public void say(int r) {
                System.out.println("! " + (r+1));
                System.out.flush();
            }
        };
        solve(q);
    }

    interface Query {
        int excuse(int u, int v);
        void say(int r);
    }

    static void solve(Query q) {
        Edge[][] G = adjB(N, E);
        int curr = 0;

        while(true) {
            int a = findNotLeaf(-1, curr, G);
            if( a == -1 ) {
                int pair = last2Node(curr, G);
                if( pair == -1 ) {
                    q.say(curr);
                } else {
                    q.say(q.excuse(curr, pair));
                }
                return;
            }

            int u = -1, v = -1;
            for (Edge e : G[a]) {
                if( e.del ) continue;

                int b = e.a == a ? e.b : e.a;
                if( u == -1 ) {
                    u = b;
                } else {
                    v = b;
                    break;
                }
            }

            if( v == -1 ) v = a;
            int w = q.excuse(u, v);

            for (Edge e : G[w]) {
                int b = e.a == w ? e.b : e.a;
                if( b == u || b == v || b == a ) {
                    e.del = true;
                }
            }
            curr = w;
        }
    }

    // ついになるものを返す or -1
    static int last2Node(int a, Edge[][] G) {
        int candidate = -1;
        for (Edge e : G[a]) {
            if( e.del ) continue;
            int b = e.a == a ? e.b : e.a;

            if( candidate == -1 ) {
                candidate = b;
            } else {
                return -1;
            }
        }
        if( candidate == -1 ) return -1;

        for (Edge e : G[candidate]){
            if( e.del ) continue;

            int b = e.a == candidate ? e.b : e.a;
            if( b != a ) return -1;
        }
        return candidate;
    }

    static int findNotLeaf(int parent, int a, Edge[][] G) {
        int cnt = 0;
        for (Edge e : G[a]) {
            if( !e.del ) {
                cnt++;
            }
        }
        if( cnt >= 2 ) return a;

        for (Edge e : G[a]) {
            if( e.del ) continue;

            int b = e.a == a ? e.b : e.a;
            if( b == parent ) continue;

            int ret = findNotLeaf(a, b, G);
            if( ret != -1 ) return ret;
        }
        return -1;
    }

    static class Edge {
        int a, b;
        boolean del;

        public Edge(int a, int b) {
            this.a = a;
            this.b = b;
        }
    }

    static Edge[][] adjB(int n, Edge[] E) {
        Edge[][] adj = new Edge[n][];
        int[] cnt = new int[n];
        for (Edge e : E) {
            cnt[e.a]++;
            cnt[e.b]++;
        }
        for (int i = 0; i < n; i++) {
            adj[i] = new Edge[cnt[i]];
        }
        for (Edge e : E) {
            adj[e.a][--cnt[e.a]] = e;
            adj[e.b][--cnt[e.b]] = e;
        }
        return adj;
    }

    @SuppressWarnings("unused")
    static class FastScanner {
        private BufferedReader reader;
        private StringTokenizer tokenizer;

        FastScanner(InputStream in) {
            reader = new BufferedReader(new InputStreamReader(in));
            tokenizer = null;
        }

        String next() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        String nextLine() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    return reader.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken("\n");
        }

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

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

        int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int[] nextIntArray(int n, int delta) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt() + delta;
            return a;
        }

        long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }
    }

    static void writeLines(int[] as) {
        PrintWriter pw = new PrintWriter(System.out);
        for (int a : as) pw.println(a);
        pw.flush();
    }

    static void writeLines(long[] as) {
        PrintWriter pw = new PrintWriter(System.out);
        for (long a : as) pw.println(a);
        pw.flush();
    }

    static void writeSingleLine(int[] as) {
        PrintWriter pw = new PrintWriter(System.out);
        for (int i = 0; i < as.length; i++) {
            if (i != 0) pw.print(" ");
            pw.print(as[i]);
        }
        pw.println();
        pw.flush();
    }

    static int max(int... as) {
        int max = Integer.MIN_VALUE;
        for (int a : as) max = Math.max(a, max);
        return max;
    }

    static int min(int... as) {
        int min = Integer.MAX_VALUE;
        for (int a : as) min = Math.min(a, min);
        return min;
    }

    static void debug(Object... args) {
        StringJoiner j = new StringJoiner(" ");
        for (Object arg : args) {
            if (arg == null) j.add("null");
            else if (arg instanceof int[]) j.add(Arrays.toString((int[]) arg));
            else if (arg instanceof long[]) j.add(Arrays.toString((long[]) arg));
            else if (arg instanceof double[]) j.add(Arrays.toString((double[]) arg));
            else if (arg instanceof Object[]) j.add(Arrays.toString((Object[]) arg));
            else j.add(arg.toString());
        }
        System.err.println(j.toString());
    }

    static void printSingleLine(int[] array) {
        PrintWriter pw = new PrintWriter(System.out);
        for (int i = 0; i < array.length; i++) {
            if (i != 0) pw.print(" ");
            pw.print(array[i]);
        }
        pw.println();
        pw.flush();
    }

    static int lowerBound(int[] array, int value) {
        int lo = 0, hi = array.length, mid;
        while (lo < hi) {
            mid = (hi + lo) / 2;
            if (array[mid] < value) lo = mid + 1;
            else hi = mid;
        }
        return lo;
    }

    static int upperBound(int[] array, int value) {
        int lo = 0, hi = array.length, mid;
        while (lo < hi) {
            mid = (hi + lo) / 2;
            if (array[mid] <= value) lo = mid + 1;
            else hi = mid;
        }
        return lo;
    }
}