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 N = 2e5 + 10;
struct edge {
  int nxt, to;
} e[N << 1];
int head[N], pos, n, tp[N], ed[N], son[N], siz[N], anc[N][20], dep[N], tdp;
void dfs1(int u, int fa) {
  int i;
  son[u] = -1;
  siz[u] = 1;
  dep[u] = dep[fa] + 1;
  for (i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa) continue;
    anc[v][0] = u;
    dfs1(v, u);
    siz[u] += siz[v];
    if (son[u] == -1 || siz[son[u]] < siz[v]) son[u] = v;
  }
}
void dfs2(int u, int fa) {
  int i;
  tp[u] = fa;
  if (son[u] == -1) {
    ed[fa] = u;
    return;
  }
  dfs2(son[u], fa);
  for (i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == anc[u][0] || v == son[u]) continue;
    dfs2(v, v);
  }
}
void add(int u, int v) {
  e[++pos] = (edge){head[u], v};
  head[u] = pos;
}
int T;
int qdp(int u) {
  int a;
  printf("d %d\n", u);
  fflush(stdout);
  scanf("%d", &a);
  if (!a) {
    printf("! %d", u);
    exit(0);
  }
  if (dep[u] - (dep[u] + tdp - a) / 2 == a) T = 1;
  return (dep[u] + tdp - a) >> 1;
}
int main() {
  scanf("%d", &n);
  int i, j;
  for (i = 2; i <= n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  for (j = 1; j <= 19; ++j)
    for (i = 1; i <= n; ++i) anc[i][j] = anc[anc[i][j - 1]][j - 1];
  printf("d 1\n");
  fflush(stdout);
  scanf("%d", &tdp);
  ++tdp;
  int u = 1;
  int mx = -1;
  while (1) {
    u = ed[u];
    int DEP = qdp(u);
    for (i = 19; i >= 0; --i)
      if (dep[anc[u][i]] >= DEP) u = anc[u][i];
    if (dep[u] == mx) {
      printf("! %d", u);
      return 0;
    }
    mx = dep[u];
    if (T) {
      printf("! %d", u);
      return 0;
    }
    printf("s %d\n", u);
    fflush(stdout);
    scanf("%d", &u);
  }
}