Problem: One day, Takahashi was given the following problem from Aoki:

* You are given a tree with N vertices and an integer K. The vertices are numbered 1 through N. The edges are represented by pairs of integers (a_i, b_i).
* For a set S of vertices in the tree, let f(S) be the minimum number of the vertices in a subtree of the given tree that contains all vertices in S.
* There are <image> ways to choose K vertices from the trees. For each of them, let S be the set of the chosen vertices, and find the sum of f(S) over all <image> ways.
* Since the answer may be extremely large, print it modulo 924844033(prime).



Since it was too easy for him, he decided to solve this problem for all K = 1,2,...,N.

Constraints

* 2 ≦ N ≦ 200,000
* 1 ≦ a_i, b_i ≦ N
* The given graph is a tree.

Input

The input is given from Standard Input in the following format:


N
a_1 b_1
a_2 b_2
:
a_{N-1} b_{N-1}


Output

Print N lines. The i-th line should contain the answer to the problem where K=i, modulo 924844033.

Examples

Input

3
1 2
2 3


Output

3
7
3


Input

4
1 2
1 3
1 4


Output

4
15
13
4


Input

7
1 2
2 3
2 4
4 5
4 6
6 7


Output

7
67
150
179
122
45
7

Solution (in CPP):

Solution:
#include<cctype>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
const int N=524288,P=924844033;
int n,lim,fac[N],ifac[N],rev[N],w[N],f[N],g[N];std::vector<int>e[N];
int read(){int x=0,c=getchar();while(isspace(c))c=getchar();while(isdigit(c))(x*=10)+=c&15,c=getchar();return x;}
void inc(int&a,int b){a+=b-P,a+=a>>31&P;}
int sub(int a,int b){return a-=b,a+=a>>31&P;}
int mul(int a,int b){return 1ll*a*b%P;}
int pow(int a,int k){int r=1;for(;k;k>>=1,a=mul(a,a))if(k&1)r=mul(a,r);return r;}
int C(int n,int m){return 1ll*fac[n]*ifac[m]%P*ifac[n-m]%P;}
void init()
{
    int p=(lim=1<<(33-__builtin_clz(n)))>>1,g=pow(5,(P-1)/lim);
    w[p]=fac[0]=1;
    for(int i=1;i<lim;++i) rev[i]=(rev[i>>1]>>1)|(i&1? p:0);
    for(int i=p+1;i<lim;++i) w[i]=mul(w[i-1],g);
    for(int i=p-1;i;--i) w[i]=w[i<<1];
    for(int i=1;i<=n;++i) fac[i]=mul(fac[i-1],i);
    ifac[n]=pow(fac[n],P-2);
    for(int i=n;i;--i) ifac[i-1]=mul(ifac[i],i);
}
int dfs(int u,int fa)
{
    int size=1,x;
    for(int v:e[u]) if(v^fa) ++f[x=dfs(v,u)],size+=x;
    return ++f[n-size],size;
}
void NTT(int*a,int f)
{
    if(!~f) std::reverse(a+1,a+lim);
    for(int i=1;i<lim;++i) if(i<rev[i]) std::swap(a[i],a[rev[i]]);
    for(int i=1;i<lim;i<<=1) for(int j=0,d=i<<1;j<lim;j+=d) for(int k=0,x;k<i;++k) x=mul(a[i+j+k],w[i+k]),a[i+j+k]=sub(a[j+k],x),inc(a[j+k],x);
    if(!~f) for(int i=0,x=P-(P-1)/lim;i<lim;++i) a[i]=mul(a[i],x);
}
int main()
{
    n=read(),init();
    for(int i=1,u,v;i<n;++i) u=read(),v=read(),e[u].push_back(v),e[v].push_back(u);
    dfs(1,0),f[0]=0;
    for(int i=1;i<=n;++i) f[i]=mul(f[i],fac[i]);
    std::reverse(f,f+n+1),memcpy(g,ifac,(n+1)<<2),NTT(f,1),NTT(g,1);
    for(int i=0;i<lim;++i) f[i]=mul(f[i],g[i]);
    NTT(f,-1),std::reverse(f,f+n+1);
    for(int i=1;i<=n;++i) printf("%d\n",sub(mul(n,C(n,i)),mul(f[i],ifac[i])));
}