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<string>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<stdio.h>
#include<algorithm>
using namespace std;

const int maxn=2000000;
const long long M=924844033;
const long long g=5;
typedef long long LL;

LL A[maxn];
LL B[maxn];

int Rev[maxn];
int N,Lg;

struct edge
{
	int obj;
	edge *Next;
} e[maxn<<1];
edge *head[maxn];
int cur=-1;

int num[maxn];
int Size[maxn];
int fa[maxn];

LL fac[maxn];
LL nfac[maxn]; 
int n;

void Add(int x,int y)
{
	cur++;
	e[cur].obj=y;
	e[cur].Next=head[x];
	head[x]=e+cur;
}

void Dfs(int node)
{
	Size[node]=1;
	for (edge *p=head[node]; p; p=p->Next)
	{
		int son=p->obj;
		if (son==fa[node]) continue;
		fa[son]=node;
		Dfs(son);
		Size[node]+=Size[son];
		num[ Size[son] ]++;
	}
	num[ n-Size[node] ]++;
}

LL Pow(LL x,LL y)
{
	if (!y) return 1;
	LL temp=Pow(x,y>>1);
	temp=temp*temp%M;
	if (y&1) temp=temp*x%M;
	return temp;
}

void DFT(LL *a,int f)
{
	for (int i=0; i<N; i++)
		if (i<Rev[i]) swap(a[i],a[ Rev[i] ]);
	
	for (int len=2; len<=N; len<<=1)
	{
		int mid=(len>>1);
		LL e=Pow(g,(M-1)/len);
		if (f==-1) e=Pow(e,M-2);
		
		for (LL *p=a; p!=a+N; p+=len)
		{
			LL wn=1;
			for (int i=0; i<mid; i++)
			{
				LL temp=wn*p[mid+i]%M;
				p[mid+i]=(p[i]-temp+M)%M;
				p[i]=(p[i]+temp)%M;
				wn=wn*e%M;
			}
		}
	}
}

void NTT()
{
	N=1,Lg=0;
	while (N<2*n+2) N<<=1,Lg++;
	for (int i=0; i<N; i++)
		for (int j=0; j<Lg; j++)
			if (i&(1<<j)) Rev[i]|=(1<<(Lg-j-1));
	
	DFT(A,1);
	DFT(B,1);
	for (int i=0; i<N; i++) A[i]=A[i]*B[i]%M;
	DFT(A,-1);
	
	LL inv=Pow(N,M-2);
	for (int i=0; i<N; i++) A[i]=A[i]*inv%M;
}

int main()
{
	
	scanf("%d",&n);
	for (int i=1; i<=n; i++) head[i]=NULL;
	for (int i=1; i<n; i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		Add(u,v);
		Add(v,u);
	}
	
	Dfs(1);
	
	fac[0]=1;
	for (LL i=1; i<=n; i++) fac[i]=fac[i-1]*i%M;
	for (int i=0; i<=n; i++) nfac[i]=Pow(fac[i],M-2);
	
	for (int i=0; i<=n; i++) A[i]=(long long)num[i]*fac[i]%M;
	for (int i=0; i<=n; i++) B[i]=nfac[i];
	for (int i=0; n-i>i; i++) swap(B[i],B[n-i]);
	
	NTT();
	
	for (int k=1; k<=n; k++)
	{
		LL ans=(long long)n*fac[n]%M;
		ans=ans*nfac[k]%M;
		ans=ans*nfac[n-k]%M;
		LL Dec=A[n+k];
		Dec=Dec*nfac[k]%M;
		ans=(ans-Dec+M)%M;
		printf("%lld\n",ans);
	}
	
	return 0;
}




















