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
//2019.1.14 12:52~16:49 PaperCloud
#include<bits/stdc++.h>
#define ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
#define MN 200005
#define mod 924844033
#define g 5
#define invg 554906420
struct edge{int to,nex;}e[MN<<1];
int N,en,hr[MN];
inline void ins(int f,int t)
{
	e[++en]=(edge){t,hr[f]};hr[f]=en;
	e[++en]=(edge){f,hr[t]};hr[t]=en;
}
int fac[MN],inv[MN],A[524288],B[524288],siz[MN];
inline int fpow(int x,int m){int r=1;for(;m;m>>=1,x=1ll*x*x%mod) if(m&1) r=1ll*r*x%mod;return r;}
inline int C(int m,int n){return 1ll*fac[m]*inv[n]%mod*1ll*inv[m-n]%mod;}
inline void init()
{
	register int i;
	for(i=fac[0]=1;i<=N;++i) fac[i]=1ll*fac[i-1]*i%mod;
	for(inv[N]=fpow(fac[N],mod-2),i=N-1;~i;--i) inv[i]=1ll*inv[i+1]*(i+1)%mod;
}
int cnt=0;
inline void dfs(int x=1,int f=0)
{
	register int i;siz[x]=1;
	for(i=hr[x];i;i=e[i].nex)if(e[i].to^f) dfs(e[i].to,x),siz[x]+=siz[e[i].to],A[siz[e[i].to]]--;
	A[N-siz[x]]--;
}
int M,di,pos[524288],invM;

inline void NTT(int *a,int type)
{
    register int i,j,p,k;
    for(i=0;i<M;++i)if(i<pos[i]) std::swap(a[i],a[pos[i]]);
    for(i=1;i<M;i<<=1)
    {
        ll wn=fpow(type>0?g:invg,(mod-1)/(i<<1));
        for(p=i<<1,j=0;j<M;j+=p) 
        {
            ll w=1;
            for(k=0;k<i;++k,w=w*wn%mod)
            {
                ll X=a[j+k],Y=w*a[j+i+k]%mod;
                a[j+k]=(X+Y)%mod;a[j+i+k]=(X-Y+mod)%mod;
            }
        }
    }
    if(type==-1) for(i=0;i<M;++i) a[i]=1ll*a[i]*invM%mod;
}
int main()
{
	N=read();
	register int i,x;
	for(i=1;i<N;++i) x=read(),ins(x,read());
	init();dfs();
	for(i=1;i< N;++i) A[i]=(A[i]+mod)%mod;A[N]=N;
	for(i=1;i<=N;++i) A[i]=1ll*A[i]*fac[i]%mod;
	for(i=0;i<=N;++i) B[i]=inv[N-i];A[0]=0;
	for(M=1,di=0;M<=N<<1;M<<=1,di++);invM=fpow(M,mod-2);
	for(i=0;i<M;++i) pos[i]=(pos[i>>1]>>1)|((i&1)<<(di-1));
	NTT(A,1);NTT(B,1);
	for(i=0;i<M;++i) A[i]=1ll*A[i]*B[i]%mod;
	NTT(A,-1);
	
	for(i=1;i<=N;++i)
	printf("%d\n",(1ll*A[N+i]*inv[i])%mod);
	return 0;
}