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.
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.Stack;

public class BigFinale {
    static final Scanner scanner = new Scanner(System.in);
    static final Node[] nodes = new Node[200001];

    public static void main(String[] args) {
        int n = scanner.nextInt();
        for (int i = 1; i <= n; i++) {
            nodes[i] = new Node(i);
        }
        for (int j = 1; j < n; j++) {
            int a = scanner.nextInt();
            int b = scanner.nextInt();
            nodes[a].children.add(nodes[b]);
            nodes[b].children.add(nodes[a]);
        }
        nodes[1].depth = 0;
        Stack<Node> stack = new Stack<>();
        stack.push(nodes[1]);
        while (!stack.isEmpty()) {
            Node node = stack.pop();
            if (node.seen) {
                node.subTreeSize = 1;
                for (Node child : node.children) {
                    if (node.parent != child) {
                        node.subTreeSize += child.subTreeSize;
                        if (node.bestChild == null || child.subTreeSize > node.bestChild.subTreeSize) {
                            node.bestChild = child;
                            node.bestDescendant = child.bestDescendant;
                        }
                    }
                }
                if (node.bestDescendant == null) {
                    node.bestDescendant = node;
                }
                /*if (node.bestChild != null) {
                    System.out.println("nodes[" + node.ix + "].bestChild = nodes[" + node.bestChild.ix + "]");
                } else {
                    System.out.println("nodes[" + node.ix + "].bestChild = null");
                }
                System.out.println("nodes[" + node.ix + "].bestDescendant = nodes[" + node.bestDescendant.ix + "]");*/
            } else {
                node.seen = true;
                stack.push(node);
                for (Node child : node.children) {
                    if (node.parent != child) {
                        child.parent = node;
                        child.depth = node.depth + 1;
                        stack.push(child);
                    }
                }
            }
        }
        int xDepth = xDist(nodes[1]);
        Node node = nodes[1];
        while (node.depth != xDepth) {
            int descDist = xDist(node.bestDescendant);
            int peakDepth = (xDepth + node.bestDescendant.depth - descDist) >> 1;
            for (; node.depth < peakDepth; node = node.bestChild);
            if (node.depth == xDepth) {
                break;
            }
            node = nextChild(node);
        }
        System.out.println("! " + node.ix);
    }

    static int xDist(Node node) {
        System.out.println("d " + node.ix);
        int res = scanner.nextInt();
        if (res == -1) {
            System.exit(0);
        }
        return res;
    }

    static Node nextChild(Node node) {
        System.out.println("s " + node.ix);
        int res = scanner.nextInt();
        if (res == -1) {
            System.exit(0);
        }
        return nodes[res];
    }

    static class Node {
        final int ix;
        int depth = -1;
        Node parent = null;
        boolean seen = false;
        List<Node> children = new ArrayList<>();
        int subTreeSize = -1;
        Node bestChild = null;
        Node bestDescendant = null;

        Node(int ix) {
            this.ix = ix;
        }
    }
}
