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<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 800000
#define MO 924844033
#define G 5
using namespace std;
typedef long long LL;
struct node
{
	int to;
	node *nxt;
}edges[MAXN*2+5];
node *ncnt=&edges[0],*Adj[MAXN+5];
int n;
LL siz[MAXN+5],fact[MAXN+5],inv[MAXN+5],f[MAXN+5];
LL a[MAXN+5],b[MAXN+5],c[MAXN+5],ans[MAXN+5];
LL PowMod(LL a,LL b)
{
	LL ret=1LL;
	while(b)
	{
		if(b&1)
			ret=1LL*ret*a%MO;
		a=1LL*a*a%MO;
		b>>=1;
	}
	return ret;
}
void prepare()
{
	fact[0]=1LL;
	for(int i=1;i<=MAXN;i++)
		fact[i]=1LL*fact[i-1]*i%MO;
	inv[MAXN]=PowMod(fact[MAXN],MO-2);
	for(int i=MAXN-1;i>=0;i--)
		inv[i]=1LL*inv[i+1]*(1LL*i+1LL)%MO;
}
void AddEdge(int u,int v)
{
	node *p=++ncnt;
	p->to=v;
	p->nxt=Adj[u];
	Adj[u]=p;
	
	node *q=++ncnt;
	q->to=u;
	q->nxt=Adj[v];
	Adj[v]=q;
}
void DFS(int u,int fa)
{
	siz[u]=1;
	for(node *p=Adj[u];p!=NULL;p=p->nxt)
	{
		int v=p->to;
		if(v==fa)
			continue;
		DFS(v,u);
		siz[u]+=siz[v];
	}
}
void NTT(LL P[],int N,int oper)
{
	for(int i=1,j=0;i<N-1;i++)
	{
		for(LL s=N;j^=s>>=1,~j&s;);
		if(i<j)
			swap(P[i],P[j]);
	}
	for(int d=0;(1<<d)<N;d++)
	{
		LL unit_p0;
		LL m=(1<<d),m2=m*2;
		unit_p0=PowMod(G,(MO-1)/m2);
		if(oper==-1)
			unit_p0=PowMod(unit_p0,MO-2);
		for(int i=0;i<N;i+=m2)
		{
			LL unit=1;
			for(int j=0;j<m;j++)
			{
				LL &P1=P[i+j+m],&P2=P[i+j];
				LL t=1LL*unit*P1%MO;
				P1=(1LL*P2-1LL*t+1LL*MO)%MO;
				P2=(1LL*P2+1LL*t)%MO;
				unit=1LL*unit*unit_p0%MO;
			}
		}
	}
	if(oper==-1)
	{
		LL Inv=PowMod(N,MO-2);
		for(int i=0;i<=N;i++)
			P[i]=1LL*P[i]*Inv%MO;
	}
}
int main()
{
//	freopen("calculate.in","r",stdin);
//	freopen("calculate.out","w",stdout);
	prepare();
	scanf("%d",&n);
	int u,v;
	for(int i=1;i<n;i++)
	{
		scanf("%d %d",&u,&v);
		AddEdge(u,v);
	}
	DFS(1,-1);
	for(int i=1;i<=n;i++)
	{
		f[siz[i]]--;
		f[n-siz[i]]--;
	}
	f[n]=n;
	int len=1;
	for(;len<=n*2;len*=2);
	for(int i=0;i<=n;i++)
		a[i]=1LL*f[i]*fact[i]%MO;
	for(int i=0;i<=n;i++)
		b[i]=inv[n-i];
	NTT(a,len,1);
	NTT(b,len,1);
	for(int i=0;i<=len;i++)
		c[i]=1LL*a[i]*b[i]%MO;
	NTT(c,len,-1);
	for(int i=1;i<=n;i++)
		ans[i]=(1LL*inv[i]*c[i+n]%MO+MO)%MO;
	for(int i=1;i<=n;i++)
		printf("%lld\n",ans[i]);
	return 0;
}