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:
/*
 ID: tommatt1
 LANG: JAVA
 TASK: 
*/
//JMJ+
import java.util.*;
import java.io.*;
public class cf1174f{
static ArrayList<Integer>[] adj;
static int[] d;
static int[] sz;
static int ori;
static int[] hv;
static int[] bot;
static BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
	public static void main(String[] args)throws IOException {
		StringTokenizer st=new StringTokenizer(bf.readLine());
		int n=Integer.parseInt(st.nextToken());
		adj=new ArrayList[n+1];
		bot=new int[n+1];
		hv=new int[n+1];
		d=new int[n+1];
		sz=new int[n+1];
		for(int i=0;i<=n;i++) {
			adj[i]=new ArrayList<>();
		}
		for(int i=1;i<n;i++) {
			st=new StringTokenizer(bf.readLine());
			int a1=Integer.parseInt(st.nextToken());
			int a2=Integer.parseInt(st.nextToken());
			adj[a1].add(a2);
			adj[a2].add(a1);
		}
		System.out.println("d "+1);
		System.out.flush();
		ori=Integer.parseInt(bf.readLine());
		if(ori==0) {
			System.out.println("! "+1);
			System.out.flush();
			return;
		}
		hld(1);
		System.out.println("! "+dfs(1));
		System.out.flush();
	}
	static int dfs(int x) throws IOException{
		for(int len=ori+d[bot[x]]-qry('d',bot[x])>>1;d[x]<len;) {
			x=hv[x];
		}
		return d[x]==ori?x:dfs(qry('s',x));
	}
	static void hld(int x) {
		sz[x]=1;
		bot[x]=x;
		for(int i:adj[x]) {
			if(sz[i]==0) {
				d[i]=d[x]+1;
				hld(i);
				sz[x]+=sz[i];
				if(sz[hv[x]]<sz[i]) {
					hv[x]=i;
					bot[x]=bot[hv[x]];
				}
			}
		}
	}
	static int qry(char c, int x) throws IOException{
		System.out.println(c+" "+x);
		System.out.flush();
		return Integer.parseInt(bf.readLine());
	}
}