1 /*
2 题意：给出一个树，根是1，每次选一个节点删除，求期望删空的步数e = a/b, 求x使得(xb = a) % 998244353.
3 题解：考虑每个节点对答案的贡献，一个节点i被删除时，对答案贡献1，而被删除的概率为1/dep[i].
4 e = sum(1/dep[i]) = a/b,
5 由 (xb=a)%P => x=a*(b^-1) % P; b^-1 是模P下的逆元
6 于是x = sum(1*(dep[i]^-1)) % P;
7 时间：2018.08.02
8 */
9
10 #include <bits/stdc++.h>
11 using namespace std;
12
13 typedef long long LL;
14 const int MAXN = 100005;
15 const LL MOD7 = 1e9+7;
16 const LL MOD717 = 7*17*(1<<23) + 1;
17
18 vector<int> g[MAXN];
19 int depth[MAXN];
20 int n;
21
22 void dfs(int u,int pre,int dep)
23 {
24 depth[u]=dep;
25 for (int v:g[u])
26 {
27 if (v==pre) continue;
28 dfs(v,u,dep+1);
29 }
30 }
31
32 LL Pow(LL a, LL b, LL P)
33 {
34 LL res=1LL;
35 LL ans=a%P;
36 while (b)
37 {
38 if (b&1) res=res*ans%P;
39 ans=ans*ans%P;
40 b>>=1;
41 }
42 return res;
43 }
44
45
46 int main()
47 {
48 // cout<<MOD717<<endl;
49 #ifndef ONLINE_JUDGE
50 // freopen("test.txt","r",stdin);
51 #endif // ONLINE_JUDGE
52 int u,v;
53 while (scanf("%d",&n)!=-1)
54 {
55 for (int i=0;i<=n;++i)
56 {
57 g[i].clear();
58 depth[i]=0;
59 }
60 for (int i=1;i<n;++i)
61 {
62 scanf("%d%d",&u,&v);
63 g[u].push_back(v);
64 g[v].push_back(u);
65 }
66 dfs(1, -1, 1);
67 LL ans=0;
68 for (int i=1;i<=n;++i)
69 {
70 ans = (ans + Pow(depth[i], MOD717-2, MOD717)) % MOD717;
71 }
72 printf("%lld\n",ans);
73 }
74 return 0;
75 }
