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 CF1174F {
	static Scanner sc = new Scanner(System.in);
	static int query(char q, int i) {
		System.out.println(q + " " + i);
		System.out.flush();
		int a = sc.nextInt();
		if (a == -1)
			System.exit(0);
		return a;
	}
	static class V {
		ArrayList<Integer> adj = new ArrayList<>();
		int p, d, n;
		boolean deleted;
	}
	static V[] vv;
	static int[] qq;
	static int cnt, dx;
	static void dfs(int p, int i, int d) {
		vv[i].p = p;
		vv[i].d = d;
		int n;
		if (d == dx)
			n = 1;
		else {
			n = 0;
			for (int j : vv[i].adj)
				if (j != p) {
					dfs(i, j, d + 1);
					n += vv[j].n;
				}
		}
		vv[i].n = n;
	}
	static void gather(int i) {
		if (vv[i].d == dx) {
			qq[cnt++] = i;
			return;
		}
		for (int j : vv[i].adj)
			if (j != vv[i].p && !vv[j].deleted && vv[j].n > 0)
				gather(j);
	}
	static Random rand = new Random();
	static int solve() {
		int i = 1;
		while (vv[i].d < dx) {
			int sum = 0, n_ = 0, j_ = -1;
			for (int j : vv[i].adj)
				if (j != vv[i].p && !vv[j].deleted && vv[j].n > 0) {
					int n = vv[j].n;
					sum += n;
					if (n_ < n) {
						n_ = n;
						j_ = j;
					}
				}
			if (n_ == sum) {
				i = j_;
				continue;
			}
			if (n_ * 2 <= sum)
				i = query('s', i);
			else {
				cnt = 0;
				gather(j_);
				i = qq[rand.nextInt(cnt)];
				for (int d = query('d', i) / 2; d > 0; d--) {
					vv[i].deleted = true;
					i = vv[i].p;
				}
			}
		}
		return i;
	}
	public static void main(String[] args) {
		int n = sc.nextInt();
		vv = new V[n + 1];
		for (int i = 1; i <= n; i++)
			vv[i] = new V();
		for (int h = 0; h < n - 1; h++) {
			int i = sc.nextInt();
			int j = sc.nextInt();
			vv[i].adj.add(j);
			vv[j].adj.add(i);
		}
		dx = query('d', 1);
		dfs(0, 1, 0);
		qq = new int[n];
		System.out.println("! " + solve());
		System.out.flush();
	}
}