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;
const int MAXN = 1e6 + 10;
int n, m;
vector<int> edge[MAXN];
int dep[MAXN], sz[MAXN], del[MAXN], f[MAXN];
int query(char ch, int u) {
  int ret;
  printf("%c %d\n", ch, u);
  fflush(stdout);
  scanf("%d", &ret);
  fflush(stdout);
  return ret;
}
void dfs1(int u, int fa) {
  if (fa == 0) dep[u] = 1;
  f[u] = fa;
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    if (v == fa) continue;
    dep[v] = dep[u] + 1;
    dfs1(v, u);
  }
}
void dfs2(int u, int fa) {
  sz[u] = 1;
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    if (v == fa || del[v] == 1) continue;
    dfs2(v, u);
    sz[u] += sz[v];
  }
}
int getCen(int u, int fa, int lim) {
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    if (v == fa || del[v] == 1) continue;
    if (sz[v] * 2 >= lim) return getCen(v, u, lim);
  }
  return u;
}
int dfs3(int u) {
  dfs2(u, 0);
  int cen = getCen(u, 0, sz[u]);
  if (dep[cen] + query('d', cen) == dep[0]) {
    if (dep[cen] == dep[0]) return cen;
    del[cen] = 1;
    return dfs3(query('s', cen));
  } else {
    del[cen] = 1;
    return dfs3(f[cen]);
  }
}
int main() {
  scanf("%d", &n);
  memset(del, 0, sizeof(del));
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  dfs1(1, 0);
  dep[0] = query('d', 1) + 1;
  int ans = dfs3(1);
  printf("! %d\n", ans);
  return 0;
}