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

public class A {

	static Scanner sc = new Scanner();
	static int log=22;
	static ArrayList<Integer> cand;
	static ArrayList<Integer>[] adj;
	static int[] par, level,up[];

	static void dfs(int u, int p) {
		up[u][0]=p;
		for(int i=1;i<log;i++)
			up[u][i]=up[up[u][i-1]][i-1];
		for (int v : adj[u])
			if (v != p) {
				par[v] = u;
				level[v] = level[u] + 1;
				dfs(v, u);
			}
	}
	static int getParent(int u,int num) {
		for(int bit=0;bit<log;bit++)
			if((num&(1<<bit))>0) {
				u=up[u][bit];
//				num-=(1<<bit);
			}
//		System.out.println(u);
		return u;
	}
	static void solve(int u, int lvl) {
		if (level[u] == lvl)
			cand.add(u);
		for (int v : adj[u])
			if (level[v] > level[u])
				solve(v, lvl);
	}

	static int query(int u, int t) throws IOException {
		System.out.println((t == 1 ? "d " : "s ") + u);
		return sc.nextInt();
		
	}
	static void answer(int u) {
		System.out.println("! "+u);
	}
	public static void main(String[] args) throws IOException {
		PrintWriter out = new PrintWriter(System.out);
		int n = sc.nextInt();
		adj = new ArrayList[n + 1];
		for (int i = 1; i <= n; i++)
			adj[i] = new ArrayList();
		for (int i = 1; i < n; i++) {
			int u = sc.nextInt(), v = sc.nextInt();
			adj[u].add(v);
			adj[v].add(u);
		}
		up=new int [n+1][log];
		par=new int [n+1];
		level=new int [n+1];
		cand = new ArrayList();
		dfs(1, 1);
		int lvl=query(1, 1);
		solve(1,lvl);
		while(true) {
			if(cand.size()==1) {
				answer(cand.get(0));
				return;
			}
			if(cand.size()==2) {
				int d=query(cand.get(0),1);
				answer(cand.get(d==0?0:1));
				return;
			}
			int u=cand.get(cand.size()/2);
			int d=query(u,1);
			if(d==0) {
				answer(u);
				break;
			}
			int lca_level=lvl-d/2;
			u=getParent(u, level[u]-lca_level);
			cand.clear();
			int v=query(u,2);
			solve(v,lvl);
			
			
		}
//		out.close();

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

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

		Scanner(String fileName) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(fileName));
		}

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

		String nextLine() throws IOException {
			return br.readLine();
		}

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

		long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}

		boolean ready() throws IOException {
			return br.ready();
		}

	}

}