Problem: This is an interactive problem.

You're given a tree consisting of n nodes, rooted at node 1. A tree is a connected graph with no cycles.

We chose a hidden node x. In order to find this node, you can ask queries of two types: 

  * d u (1 ≤ u ≤ n). We will answer with the distance between nodes u and x. The distance between two nodes is the number of edges in the shortest path between them. 
  * s u (1 ≤ u ≤ n). We will answer with the second node on the path from u to x. However, there's a plot twist. If u is not an ancestor of x, you'll receive "Wrong answer" verdict! 



Node a is called an ancestor of node b if a ≠ b and the shortest path from node 1 to node b passes through node a. Note that in this problem a node is not an ancestor of itself.

Can you find x in no more than 36 queries? The hidden node is fixed in each test beforehand and does not depend on your queries.

Input

The first line contains the integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of nodes in the tree.

Each of the next n-1 lines contains two space-separated integers u and v (1 ≤ u,v ≤ n) that mean there's an edge between nodes u and v. It's guaranteed that the given graph is a tree.

Output

To print the answer, print "! x" (without quotes).

Interaction

To ask a question, print it in one of the formats above: 

  * d u (1 ≤ u ≤ n), or 
  * s u (1 ≤ u ≤ n). 



After each question, you should read the answer: either the distance or the second vertex on the path, as mentioned in the legend. 

If we answer with -1 instead of a valid answer, that means you exceeded the number of queries, made an invalid query, or violated the condition in the second type of queries. Exit immediately after receiving -1 and you will see Wrong answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.

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 the documentation for other languages.



Hacks:

The first line should contain two integers n and x (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ x ≤ n).

Each of the next n-1 lines should contain two integers u and v (1 ≤ u,v ≤ n) that mean there is an edge between nodes u and v. The edges must form a tree.

Example

Input


5
1 2
1 3
3 4
3 5
3
5

Output


d 2
s 3
! 5

Note

In the first example, the hidden node is node 5.

<image>

We first ask about the distance between node x and node 2. The answer is 3, so node x is either 4 or 5. We then ask about the second node in the path from node 3 to node x. Note here that node 3 is an ancestor of node 5. We receive node 5 as the answer. Finally, we report that the hidden node is node 5.

Solution (in JAVA):

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

import java.util.*;
import java.io.*;

public class F
{
    int N , L[] , P[];
    int [] size ;
    int [] label ;
    int [][] adjList ;
    Scanner sc = new Scanner(System.in);
    PrintWriter out = new PrintWriter(System.out);
    int ask(char s , int node) throws Exception
    {
        out.println(s+" "+(node + 1));
        out.flush();
        return sc.nextInt() ;
    }
    void answer(int node)
    {
        out.println("! "+(node + 1));
        out.flush();
    }
    void dfs(int u , int p)
    {
        P[u] = p ; 
        for(int v : adjList[u])
            if(v != p) {
                L[v] = L[u] + 1;
                dfs(v , u) ;
            }
    }
    
    int getSize(int u , int p)
    {
        int ans = 1 ;
        for(int v : adjList[u])
            if(v != p && label[v] == 0)
                ans += getSize(v , u) ;
        return size[u] = ans;
    }
    int getCentroid(int u , int p , int n)
    {
        for(int v : adjList[u])
            if(v != p && label[v] == 0 && size[v] * 2 > n )
                return getCentroid(v , u , n) ;
        return u ;
    }
    void  decompose(int u , int lb , int dist ) throws Exception
    {
        int centorid = getCentroid(u , -1 , getSize(u , -1)) ;
        label[centorid] = lb ;
        int U_C = L[centorid] ; // dist (u -> c)
        int C_X =  ask('d' , centorid) ; // dist (c -> x)
        if(C_X == 0)
        {
            answer(centorid);
            return;
        }
        if(U_C + C_X == dist)
            decompose(ask('s' , centorid) - 1, lb + 1 , dist);
        else
            decompose(P[centorid] , lb + 1 , dist);
    }
    void main() throws Exception
    {
        N = sc.nextInt() ;
        adjList = new int [N][] ;
        int [] x = new int [N - 1] , y  = new int [N - 1] , deg = new int [N];
        label = new int [N] ;
        size = new int [N] ;
        P = new int [N] ; 
        for(int i = 0 ;i < N - 1 ; i++)
        {
            deg[x[i] = sc.nextInt() - 1]++ ;
            deg[y[i] = sc.nextInt() - 1]++ ;
        }
        for(int i = 0 ; i < N ;i++) adjList[i] = new int [deg[i]] ;
        for(int i = 0 ;i < N - 1;i++)
        {
            adjList[x[i]][--deg[x[i]]] = y[i] ;
            adjList[y[i]][--deg[y[i]]] = x[i] ;
        }
        L = new int [N] ;
        dfs(0 , -1);
        decompose(0 , 1 , ask('d' , 0));
        out.flush();
        out.close();
    }

    class Scanner
    {
        BufferedReader br;
        StringTokenizer st;

        Scanner(InputStream in)
        {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() throws Exception
        {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

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

        double nextDouble() throws Exception { return Double.parseDouble(next());}
    }
    public static void main (String [] args) throws Exception {(new F()).main();}

}