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>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<math.h>
using namespace std;

const int N=2e5+5;
const int M=5e5+5;
const int T=1e6+5;
const int mod=924844033;
const int sqr=5;
const int sqrinv=554906420;

int n,u,v,siz[N],limit,lg;
int E,ot[M],ne[M],g[N];
int fac[N],inv[N],infac[N];
int a[T],b[T],r[T];

int F_p(int x,int y)
{
	int bas=x,ret=1;
	while(y)
	{
		if(y&1) ret=1LL*ret*bas%mod;
		bas=1LL*bas*bas%mod;
		y>>=1;
	}
	return ret;
}

void add(int u,int v)
{
	ot[++E]=v;
	ne[E]=g[u];
	g[u]=E;
}

void init()
{
	int i;
	scanf("%d",&n);
	for(i=1;i<n;++i)
	{
		scanf("%d%d",&u,&v);
		add(u,v),add(v,u);
	}
}

void dfs(int x,int fath)
{
	siz[x]=1;
	int p,to;
	for(p=g[x];p;p=ne[p])
	{
		to=ot[p];
		if(to==fath) continue;
		dfs(to,x);
		siz[x]+=siz[to];
		++a[siz[to]];
	}
	++a[n-siz[x]];
}

int solve_complex(int x,int tp)
{
	if(~tp) return F_p(sqr,x);
	else return F_p(sqrinv,x);
}

void DFT(int *A,int tp)
{
	int i;
	for(i=0;i<limit;++i)
		if(i<r[i])
			swap(A[i],A[r[i]]);
	
	int len,j,k;
	for(len=1;len<limit;len<<=1)
	{
		int wn=solve_complex((mod-1)/(2*len),tp);
		for(j=0;j<limit;j+=len<<1)
		{
			int w=1;
			for(k=0;k<len;++k,w=1LL*w*wn%mod)
			{
				int x=A[j+k];
				int y=1LL*w*A[j+k+len]%mod;
				A[j+k]=x+y,A[j+k+len]=x-y;
				if(A[j+k]>=mod) A[j+k]-=mod;
				if(A[j+k+len]<0) A[j+k+len]+=mod;
			}
		}
	}
}

void prework()
{
	int i;dfs(1,0);
	fac[0]=1;for(i=1;i<=n;++i) fac[i]=1LL*fac[i-1]*i%mod;
	inv[1]=1;for(i=2;i<=n;++i) inv[i]=1LL*inv[mod%i]*(mod-mod/i)%mod;
	infac[0]=1;for(i=1;i<=n;++i) infac[i]=1LL*infac[i-1]*inv[i]%mod;
}

void work()
{
	int i;a[0]=0;
	for(i=0;i<=n;++i) a[i]=1LL*a[i]*fac[i]%mod;
	for(i=0;i<=n;++i) b[i]=infac[n-i];
	
	limit=1;
	while(limit<=2*n) limit<<=1,++lg;
	for(i=1;i<limit;++i) r[i]=(r[i>>1]>>1)|((i&1)<<(lg-1));
	
	DFT(a,1),DFT(b,1);
	for(i=0;i<limit;++i) a[i]=1LL*a[i]*b[i]%mod;
	DFT(a,-1);
	
	int INV=F_p(limit,mod-2);
	for(i=1;i<=n;++i) a[i+n]=1LL*a[i+n]*INV%mod*infac[i]%mod;
	for(i=1;i<=n;++i) a[i]=1LL*fac[n]*infac[i]%mod*infac[n-i]%mod*n%mod;
	for(i=1;i<=n;++i)
	{
		a[i]-=a[i+n];
		if(a[i]<0) a[i]+=mod;
	}
	for(i=1;i<=n;++i) printf("%d\n",a[i]);
}

int main()
{
	init();prework();work();
	return 0;
}