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 java.util.*;

public class finale {
	public static void main(String[] args) {
		scan=new Scanner(System.in);
		int n=scan.nextInt();
		a=new ArrayList[n];
		for(int i=0;i<n;i++) a[i]=new ArrayList<>();
		for(int i=0;i<n-1;i++) {
			int u=scan.nextInt()-1, v=scan.nextInt()-1;
			a[u].add(v);
			a[v].add(u);
		}
		System.out.println("d "+1);
		System.out.flush();
		depthx=scan.nextInt();
		size=new int[n];
		dead=new boolean[n];
		h=new int[n];
		p=new int[n];
		fill(0,-1);
		if(abort) return;
		go(0);
	}
	public static void fill(int at, int prev) {
		p[at]=prev;
		if(prev!=-1) h[at]=h[prev]+1;
		for(int nxt:a[at]) {
			if(nxt==prev) continue;
			fill(nxt,at);
		}
	}
	static int depthx;
	static int res;
	static Scanner scan;
	static boolean[] dead;
	static ArrayList<Integer>[] a;
	static int[] size;
	public static void go(int at) {
		if(abort) return;
		int sz=updsize(at,-1);
		if(sz==1) {
			abort=true;
			System.out.println("! "+(at+1));
			System.out.flush();
			return;
		}
		int centroid=centroid(at,-1,sz);
		dead[centroid]=true;
		
		System.out.println("d "+(centroid+1));
		System.out.flush();
		int x=scan.nextInt();
		if(x==-1) {
			abort=true;
			return;
		}
		//dist=0
		if(x==0) {
			abort=true;
			System.out.println("! "+(centroid+1));
			System.out.flush();
			return;
		}
		//is centroid ancestor of node x?
		if(h[centroid]+x==depthx) {
			//it is
			System.out.println("s "+(centroid+1));
			System.out.flush();
			int sec=scan.nextInt();
			if(x==-1) {
				abort=true;
				return;
			}
			go(sec-1);
		}
		else {
			//it's not so it has to be in centroid's parent
			go(p[centroid]);
		}
	}
	static int[] h,p;
	public static int centroid(int at, int prev, int sz) {
		for(int nxt:a[at]) {
			if(nxt==prev||dead[nxt]||size[nxt]<sz/2) continue;
			return centroid(nxt,at,sz);
		}
		return at;
	}
	public static int updsize(int at, int prev) {
		size[at]=1;
		for(int nxt:a[at]) {
			if(nxt==prev||dead[nxt]) continue;
			size[at]+=updsize(nxt,at);
		}
		return size[at];
	}
	static boolean abort=false;
}