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>
const int maxn = 200005;
int head[maxn], next[maxn * 2], to[maxn * 2], lb;
int n, dep[maxn], siz[maxn], fa[maxn], heavyBottom[maxn], son[maxn];
inline void ist(int aa, int ss) {
  to[lb] = ss;
  next[lb] = head[aa];
  head[aa] = lb;
  ++lb;
}
void dfs(int r, int p) {
  fa[r] = p;
  dep[r] = dep[p] + 1;
  siz[r] = 1;
  for (int j = head[r]; j != -1; j = next[j]) {
    if (to[j] == p) {
      continue;
    }
    dfs(to[j], r);
    siz[r] += siz[to[j]];
    if (siz[son[r]] < siz[to[j]]) {
      son[r] = to[j];
    }
  }
  if (son[r] == 0) {
    heavyBottom[r] = r;
  } else {
    heavyBottom[r] = heavyBottom[son[r]];
  }
}
int main() {
  memset(next, -1, sizeof next);
  memset(head, -1, sizeof head);
  scanf("%d", &n);
  int t1, t2;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &t1, &t2);
    ist(t1, t2);
    ist(t2, t1);
  }
  dep[0] = -1;
  dfs(1, 0);
  int depx;
  printf("d %d\n", 1);
  fflush(stdout);
  scanf("%d", &depx);
  int u = 1, v, dis;
  int ans = -666;
  while (true) {
    v = heavyBottom[u];
    printf("d %d\n", v);
    fflush(stdout);
    scanf("%d", &dis);
    if (dis == 0) {
      ans = v;
      break;
    }
    dis = (dep[v] + depx - dis) / 2;
    int tm = dis - dep[u];
    for (int i = 0; i < tm; ++i) {
      u = son[u];
    }
    if (dis == depx) {
      ans = u;
      break;
    }
    printf("s %d\n", u);
    fflush(stdout);
    scanf("%d", &u);
  }
  printf("! %d\n", ans);
  return 0;
}