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 <fstream>
#include <algorithm>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;
#define mmst(a, b) memset(a, b, sizeof(a))
#define mmcp(a, b) memcpy(a, b, sizeof(b))

typedef long long LL;

const int N=800400;
const LL mo=924844033;

void read(int &hy)
{
	hy=0;
	char cc=getchar();
	while(cc<'0'||cc>'9')
	cc=getchar();
	while(cc>='0'&&cc<='9')
	{
		hy=(hy<<3)+(hy<<1)+cc-'0';
		cc=getchar();
	}
}

int nn;
int s[N];
int to[N],nex[N],head[N],cnt;
int n,rev[N];
LL num[N],I[N],jc[N],Ijc[N],ans[N];

void add(int u,int v)
{
	to[++cnt]=v;
	nex[cnt]=head[u];
	head[u]=cnt;
}

void dfs(int x)
{
	s[x]=1;
	for(int h=head[x];h;h=nex[h])
	if(!s[to[h]])
	{
		dfs(to[h]);
		s[x]+=s[to[h]];
		num[s[to[h]]]++;
		num[nn-s[to[h]]]++;
	}
}

LL cheng(LL a,LL b)
{
	LL res=1;
	for(;b;b>>=1,a=a*a%mo)
	if(b&1)
	res=res*a%mo;
	return res;
}

void Init(int lim)
{
	int k=-1;
	n=1;
	while(n<lim)
	n<<=1,k++;
	for(int i=1;i<n;i++)
	rev[i]=(rev[i>>1]>>1) | ((i&1)<<k); 
	
	I[1]=1;
	Ijc[0]=jc[0]=1;
	
	for(int i=2;i<=n;i++)
	I[i]=(mo-mo/i)*I[mo%i]%mo;
	for(int i=1;i<=n;i++)
	jc[i]=jc[i-1]*i%mo,Ijc[i]=Ijc[i-1]*I[i]%mo;
}

void ntt(LL *a,int ops)
{
	for(int i=0;i<n;i++)
	if(i<rev[i])
	swap(a[i],a[rev[i]]);
	
	for(int l=2;l<=n;l<<=1)
	{
		int m=l>>1;
		LL wn=(ops) ? cheng(5,(mo-1)/l) : cheng(5,mo-1-(mo-1)/l);
		for(int i=0;i<n;i+=l)
		{
			LL w=1;
			for(int k=0;k<m;k++)
			{
				LL t=a[i+k+m]*w%mo;
				a[i+k+m]=(a[i+k]-t+mo)%mo;
				a[i+k]=(a[i+k]+t)%mo;
				w=w*wn%mo;
			}
		}
	}
	if(!ops)
	for(int i=0;i<n;i++)
	a[i]=a[i]*I[n]%mo;
}

int main()
{
	cin>>nn;
	for(int i=1;i<nn;i++)
	{
		int u,v;
		read(u);
		read(v);
		add(u,v);
		add(v,u);
	}
	
	dfs(1);
	for(int i=1;i<nn;i++)
	num[i]=mo-num[i];
	num[nn]=nn;
	
	Init(2*nn+1);
	
	for(int i=0;i<=nn;i++)
	num[i]=num[i]*jc[i]%mo;
	
	for(int i=1;i<=nn;i++)
	ans[i]=Ijc[nn-i];
	
	ntt(num,1);	
	ntt(ans,1);
	
	for(int i=0;i<n;i++)
	ans[i]=num[i]*ans[i]%mo;
	ntt(ans,0);
	
	for(int i=1;i<=nn;i++)
	printf("%I64d\n",ans[nn+i]*Ijc[i]%mo);

	return 0;
}
