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.*;
import java.io.*;
import static java.lang.Math.*;
public class Main {
    static final long MOD = 1_000_000_007, INF = 1_000_000_000_000_000_000L;
    static final int INf = 1_000_000_000;
    static FastReader reader;
    static PrintWriter writer;
    public static void main(String[] args) {
        Thread t = new Thread(null, new O(), "Integer.MAX_VALUE", 100000000);
        t.start();
    }
    static class O implements Runnable {
        public void run() {
            try {
                magic();
            }
            catch (Exception e) {
                e.printStackTrace();
                System.exit(1);
            }
        }
    }
    static class FastReader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public FastReader() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public FastReader(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String readLine() throws IOException {
            byte[] buf = new byte[1000000];
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }

        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (c == '.') while ((c = read()) >= '0' && c <= '9') ret += (c - '0') / (div *= 10);
            if (neg) return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1) buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead) fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException {
            if (din == null) return;
            din.close();
        }
    }
    static ArrayList<Integer> adj[];
    static int lev[], subtree_size[], heavyChild[];
    static void magic() throws IOException {
        reader = new FastReader();
        writer = new PrintWriter(System.out, true);
        int n = reader.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 = reader.nextInt(), v = reader.nextInt();
            adj[u].add(v);
            adj[v].add(u);
        }
        lev = new int[n+1];
        subtree_size = new int[n+1];
        heavyChild = new int[n+1];
        dfs(1,1,0);
        int i_know_iske_subtree_me_hai = 1;
        int l = query_distance(1);
        while(true) {
            ArrayList<Integer> heavy_path = getHeavyPath(i_know_iske_subtree_me_hai);
            int last_node_in_the_path = heavy_path.get(heavy_path.size() - 1);
            int d = query_distance(last_node_in_the_path);
            int depth_of_last_common_node = (l + lev[last_node_in_the_path] - d)/2;
            int last_common_node = heavy_path.get(depth_of_last_common_node - lev[i_know_iske_subtree_me_hai]);
            if(depth_of_last_common_node==l) {
                writer.println("! "+last_common_node);
                System.exit(0);
            }
            i_know_iske_subtree_me_hai = query_second(last_common_node);
        }
    }
    static void dfs(int v, int par, int l) {
        subtree_size[v] = 1;
        lev[v] = l;
        int heavy_child = -1, heavy_child_subtree_size = 0;
        for(int e : adj[v]) {
            if(e!=par) {
                dfs(e, v, l+1);
                subtree_size[v]+=subtree_size[e];
                if(subtree_size[e]>heavy_child_subtree_size) {
                    heavy_child = e;
                    heavy_child_subtree_size = subtree_size[e];
                }
            }
        }
        heavyChild[v] = heavy_child;
    }
    static ArrayList<Integer> getHeavyPath(int v) {
        ArrayList<Integer> list = new ArrayList<>();
        while(v!=-1) {
            list.add(v);
            v = heavyChild[v];
        }
        return list;
    }
    static int query_distance(int from) throws IOException {
        writer.println("d "+from);
        writer.flush();
        return reader.nextInt();
    }
    static int query_second(int from) throws IOException {
        writer.println("s "+from);
        writer.flush();
        return reader.nextInt();
    }
}