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
#include<iostream>
#include<cstdio>
#define MN 530000
#define mod 924844033
using namespace std;
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*10+ch-'0';ch=getchar();}
	return x*f;
}
int n,head[MN+5],cnt=0,s[MN+5],size[MN+5],p[MN+5],inv[MN+5],B[MN+5],A[MN+5],w[2][MN+5],N,Inv;
struct edge{int to,next;}e[MN*2+5];
inline void ins(int f,int t)
{
	e[++cnt]=(edge){t,head[f]};head[f]=cnt;
	e[++cnt]=(edge){f,head[t]};head[t]=cnt;
}
inline int pow(int x,int k)
{
	int sum=1;
	for(;k;k>>=1,x=1LL*x*x%mod) if(k&1) sum=1LL*sum*x%mod;
	return sum;	
}
void Pre(int x,int fa)
{
	size[x]=1;
	for(int i=head[x];i;i=e[i].next) if(e[i].to!=fa)
		Pre(e[i].to,x),++s[size[e[i].to]],size[x]+=size[e[i].to];
	if(size[x]<n) ++s[n-size[x]];
}
void FFT(int*x,int r)
{
	for(int i=0,j=0;i<N;++i)
	{
		if(i>j) swap(x[i],x[j]);
		for(int k=N>>1;(j^=k)<k;k>>=1);	
	}
	for(int i=2;i<=N;i<<=1)for(int j=0;j<N;j+=i)for(int k=0;k<i>>1;++k)
	{
		int t=1LL*x[j+k+(i>>1)]*w[r][N/i*k]%mod;
		x[j+k+(i>>1)]=(x[j+k]-t+mod)%mod;
		x[j+k]=(x[j+k]+t)%mod;	
	}
	if(r)for(int i=0;i<N;++i)x[i]=1LL*x[i]*Inv%mod;
}
inline int C(int n,int m){return 1LL*p[n]*inv[m]%mod*inv[n-m]%mod;}
int main()
{
	n=read();p[0]=p[1]=inv[0]=inv[1]=1;
	for(int i=1;i<n;++i) ins(read(),read());
	Pre(1,0);
	for(int i=2;i<=n;++i) p[i]=1LL*p[i-1]*i%mod,inv[i]=1LL*(mod-mod/i)*inv[mod%i]%mod;
	for(int i=2;i<=n;++i) inv[i]=1LL*inv[i-1]*inv[i]%mod;
	for(N=1;N<=n;N<<=1);N<<=1;int g=pow(5,(mod-1)/N);Inv=pow(N,mod-2);
	for(int i=0,j=1;i<=N;++i,j=1LL*j*g%mod) w[0][i]=w[1][N-i]=j;
	for(int i=0;i<=n;++i) B[i]=inv[i],A[n-i]=1LL*s[i]*p[i]%mod;
	FFT(A,0);FFT(B,0);
	for(int i=0;i<N;++i) A[i]=1LL*A[i]*B[i]%mod;
	FFT(A,1);
	printf("%d\n",n);
	for(int i=2;i<=n;++i) printf("%d\n",((1LL*C(n,i)*n-1LL*A[n-i]*inv[i])%mod+mod)%mod);
	return 0;
}

