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 javax.print.DocFlavor;
import javax.swing.text.html.HTMLDocument;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.Inet4Address;
import java.nio.charset.IllegalCharsetNameException;
import java.sql.SQLOutput;
import java.util.*;
import java.util.concurrent.locks.ReentrantReadWriteLock;


public class Main {
    static int[] parent;
    static LinkedList<Integer>[] graph;


    public static void main(String[] args) throws IOException{
        Reader.init(System.in);
        int t = 1;
        while (t-->0) {
            solve();
        }
    }

    static TreeSet<Integer>[] g;
    static boolean v[];
    static int[] get_leaf(){
        int[] lol = new int[]{0,0};
        int c = 0;
        for (int i =0 ; i < g.length ; i++){
            if (g[i].size()==1){
                lol[c] = i;
                c++;
            }
            if (c==2){
                break;
            }
        }
        return lol;
    }
    static void remove(int n1 ,int  n2){
        g[n1] = new TreeSet<>();
        g[n2] = new TreeSet<>();
        v[n1] = true;
        v[n2] = true;
        for (int i = 0 ; i < g.length ; i++){
            if (g[i].contains(n1)){
                g[i].remove(n1);
            }
            if (g[i].contains(n2)){
                g[i].remove(n2);
            }
        }
    }
    static void solve()throws IOException {
        int n = Reader.nextInt();
        g = new TreeSet[n+1];
        v = new boolean[n+1];
        for (int i =0 ; i < n+1 ; i++){
            g[i] =new TreeSet<>();
        }
        for (int i = 0 ; i < n-1 ; i++){
            int n1 = Reader.nextInt();
            int n2 = Reader.nextInt();
            g[n1].add(n2);
            g[n2].add(n1);
        }
        for (int i = 0; i < n/2 ; i++){
            int[] lol = get_leaf();
            int n1 = lol[0];
            int n2 = lol[1];
            System.out.println("? " + n1 + " " + n2);
            System.out.flush();
            int num = Reader.nextInt();
            if (num==n1){
                System.out.println("! " + n1);
                System.exit(0);
            }
            if (num==n2){
                System.out.println("! " + n2);
                System.exit(0);
            }
            remove(n1 , n2);
        }
        for (int i = 1 ; i <= n ; i++){
            if (!v[i]){
                System.out.println("! " + i);
            }
        }

    }








    static int next(long[] arr, long target)
    {
        int start = 0, end = arr.length - 1;

        int ans = -1;
        while (start <= end) {
            int mid = (start + end) / 2;

            // Move to right side if target is
            // greater.
            if (arr[mid] <= target) {
                start = mid + 1;
            }

            // Move left side.
            else {
                ans = mid;
                end = mid - 1;
            }
        }
        return ans;
    }

    static int find(int x) {
        return parent[x] == x ? x : (parent[x] = find(parent[x]));
    }
    static boolean merge(int x, int y) {
        x = find(x);
        y = find(y);
        if (x==y){
            return false;
        }
        parent[x] = y;
        return true;
    }







    public static void sortbyColumn(int arr[][], int col)
    {
        // Using built-in sort function Arrays.sort
        Arrays.sort(arr, new Comparator<int[]>() {

            @Override
            // Compare values according to columns
            public int compare(final int[] entry1,
                               final int[] entry2) {

                // To sort in descending order revert
                // the '>' Operator
                if (entry1[col] > entry2[col])
                    return 1;
                else
                    return -1;
            }
        });  // End of function call sort().
    }


    static class Node{
        long a;
        int i;
        Node(long a, int i){
            this.a = a;
            this.i = i;
        }
    }

}


class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;

    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }
    /** get next word */
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                    reader.readLine() );
        }
        return tokenizer.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }

    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
}
class MergeSort
{
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];

        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

    /* A utility function to print array of size n */
    static void printArray(int arr[])
    {
        int n = arr.length;
        for (int i=0; i<n; ++i)
            System.out.print(arr[i] + " ");
        System.out.println();
    }

    // Driver method

}

class Node implements Comparable<Node>{
    int a;
    int b;
    Node (int a , int b){
        this.a = a;
        this.b = b;
    }



    public int compareTo(Node o) {
        if ((this.a%2) == (o.a%2)){
            return (this.b - o.b);
        }
        else{
            return this.a - o.a;
        }
    }
}