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;
void qmax(int &x, int y) {
  if (x < y) x = y;
}
void qmin(int &x, int y) {
  if (x > y) x = y;
}
inline long long read() {
  char s;
  long long k = 0, base = 1;
  while ((s = getchar()) != '-' && s != EOF && !(isdigit(s)))
    ;
  if (s == EOF) exit(0);
  if (s == '-') base = -1, s = getchar();
  while (isdigit(s)) {
    k = k * 10 + (s ^ '0');
    s = getchar();
  }
  return k * base;
}
inline void write(int x) {
  static char cnt, num[15];
  cnt = 0;
  if (!x) {
    printf("0");
    return;
  }
  for (; x; x /= 10) num[++cnt] = x % 10;
  for (; cnt; putchar(num[cnt--] + 48))
    ;
}
int D(int x) {
  printf("d %d\n", x);
  fflush(stdout);
  scanf("%d", &x);
  return x;
}
int S(int x) {
  printf("s %d\n", x);
  fflush(stdout);
  scanf("%d", &x);
  return x;
}
void Return(int x) {
  printf("! %d\n", x);
  exit(0);
}
const int maxn = 2e5 + 10;
int n;
int po[maxn], to[maxn << 1], ne[maxn << 1], id;
void add(int x, int y) {
  id++;
  to[id] = y, ne[id] = po[x], po[x] = id;
}
int fa[maxn], dfn[maxn], sz[maxn], son[maxn], top[maxn], lst[maxn], dt;
int dep[maxn];
int f[maxn][19];
void init() {
  for (int i = 1; i <= n; i++) f[i][0] = fa[i];
  for (int j = 1; j < 19; j++)
    for (int i = 1; i <= n; i++) f[i][j] = f[f[i][j - 1]][j - 1];
}
void dfs1(int x) {
  sz[x] = 1;
  dep[x] = dep[fa[x]] + 1;
  for (int i = po[x]; i; i = ne[i])
    if (to[i] != fa[x]) {
      fa[to[i]] = x;
      dfs1(to[i]);
      sz[x] += sz[to[i]];
      if (sz[to[i]] > sz[son[x]]) son[x] = to[i];
    }
}
void dfs(int x, int y) {
  dfn[x] = ++dt;
  top[x] = y;
  lst[top[x]] = x;
  if (!son[x]) return;
  dfs(son[x], y);
  for (int i = po[x]; i; i = ne[i])
    if (to[i] != son[x] && to[i] != fa[x]) dfs(to[i], to[i]);
}
int Dis;
int Find(int x, int y) {
  for (int i = 0; i < 19; i++)
    if (y & (1 << i)) x = f[x][i];
  return x;
}
void work(int x, int y) {
  if (!y) Return(x);
  int z = lst[top[x]], dis;
  dis = D(z);
  if (y - dis == dep[z] - dep[x]) {
    if (!dis) Return(z);
    z = S(z);
    work(z, dis - 1);
    return;
  }
  int p = (dis + (dep[z] - dep[x]) - y) / 2;
  z = Find(z, p);
  if (dis == p) Return(z);
  z = S(z);
  work(z, dis - p - 1);
}
int main() {
  n = read();
  for (int i = 1, x, y; i < n; i++) {
    x = read(), y = read();
    add(x, y);
    add(y, x);
  }
  dfs1(1);
  dfs(1, 1);
  init();
  Dis = D(1);
  work(1, Dis);
  return 0;
}