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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> G[200005];
int st[200005];
int vis[200005];
int totalSize, distVCent;
int D[200005];
int query(char op, int v) {
  int ret;
  printf("%c %d\n", op, v);
  fflush(stdout);
  scanf("%d", &ret);
  return ret;
}
void ans(int v) {
  printf("! %d\n", v);
  exit(0);
}
void calc_subtree(int v, int p) {
  st[v] = 1;
  for (int x : G[v]) {
    if (x == p or vis[x]) continue;
    calc_subtree(x, v);
    st[v] += st[x];
  }
}
int find_centroid(int v, int p) {
  for (int x : G[v]) {
    if (x == p or vis[x]) continue;
    if (st[x] * 2 >= totalSize) return find_centroid(x, v);
  }
  return v;
}
void dist(int cur, int destiny, int p, int d) {
  if (cur == destiny) distVCent = d;
  for (int x : G[cur]) {
    if (x == p or vis[x]) continue;
    dist(x, destiny, cur, d + 1);
  }
}
void centroid(int v) {
  int distRoot = (D[v] == -1) ? query('d', v) : D[v];
  D[v] = distRoot;
  if (distRoot == 0) {
    ans(v);
  }
  calc_subtree(v, 0);
  totalSize = st[v];
  if (totalSize == 1) {
    ans(v);
  }
  int cent = find_centroid(v, 0);
  int distCentroid = D[cent] == -1 ? query('d', cent) : D[cent];
  D[cent] = distCentroid;
  if (distCentroid == 0) {
    ans(cent);
  }
  dist(v, cent, 0, 0);
  vis[cent] = 1;
  if (distCentroid > distRoot) {
    centroid(v);
  } else if (distVCent + distCentroid == distRoot) {
    int nxt = query('s', cent);
    D[nxt] = distCentroid - 1;
    centroid(nxt);
  } else {
    centroid(v);
  }
}
int main() {
  int n, a, b;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  memset(D, -1, sizeof D);
  centroid(1);
  return 0;
}