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<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<complex>
#include<iostream>
#include<algorithm>
#define N 200001
#define LL long long
#define add_edge(u,v) nxt[++cnt]=head[u],head[u]=cnt,to[cnt]=v
#define open(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
char ch;bool fs;void re(int& x)
{
	while(ch=getchar(),ch<33);
	if(ch=='-')fs=1,x=0;else fs=0,x=ch-48;
	while(ch=getchar(),ch>33)x=x*10+ch-48;
	if(fs)x=-x;	
}
using namespace std;
const int mod=924844033;
int n,siz[N],sum[N],L,R[1<<19],a[1<<19],b[1<<19],fac[N],inv[N],ifac[N];
int cnt,head[N],nxt[N+N],to[N+N];
int mi(int a,int b)
{
	int res=1;
	while(b)
	{
		if(b&1)res=1ll*res*a%mod;
		a=1ll*a*a%mod;
		b>>=1;
	}
	return res;
}
int add(int x)
{
	return x<mod?x:x-mod;
}
int dec(int x)
{
	return x<0?x+mod:x;
}
void fft(int x[],int n,int f)
{
	for(int i=0;i<n;++i)
		if(i<R[i])
			swap(x[i],x[R[i]]);
	for(int i=1;i<n;i<<=1)
	{
		int wn=mi(5,(mod-1)/(i<<1));
		if(!~f)wn=mi(wn,mod-2);
		for(int j=0;j<n;j+=i<<1)
			for(int k=0,w=1;k<i;++k,w=1ll*w*wn%mod)
			{
				int l=x[j+k],r=1ll*x[i+j+k]*w%mod;
				x[j+k]=add(l+r);x[i+j+k]=dec(l-r);
			}
	}
	if(!~f)
	{
		int inv=mi(n,mod-2);
		for(int i=0;i<n;++i)
			x[i]=1ll*x[i]*inv%mod;
	}
}
void dfs(int u,int fa)
{
	siz[u]=1;
	for(int i=head[u],v;i;i=nxt[i])
		if((v=to[i]) != fa)
		{
			dfs(v,u);
			siz[u]+=siz[v];
		}
	if(u!=1)
		--sum[siz[u]],--sum[n-siz[u]];
}
int main()
{
	re(n);
	for(int i=1,u,v;i<n;++i)
	{
		re(u),re(v);
		add_edge(u,v);
		add_edge(v,u);
	}
	fac[0]=inv[0]=inv[1]=ifac[0]=1;
	for(int i=1;i<=n;++i)
	{
		if(i>1)inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
		fac[i]=1ll*i*fac[i-1]%mod;
		ifac[i]=1ll*inv[i]*ifac[i-1]%mod;
	}
	dfs(1,0);
	sum[n]=n;
	for(int i=0;i<=n;++i)	
		a[i]=(1ll*fac[n-i]*sum[n-i]%mod+mod)%mod,
		b[i]=ifac[i];
	int nn;
	for(nn=1;nn<=n+n;nn<<=1)++L;--L;
	for(int i=0;i<nn;++i)R[i]=(R[i>>1]>>1)|((i&1)<<L);
	fft(a,nn,1);fft(b,nn,1);
	for(int i=0;i<nn;++i)
		a[i]=1ll*a[i]*b[i]%mod;
	fft(a,nn,-1);
	for(int i=1;i<=n;++i)
		printf("%lld\n",1ll*a[n-i]*ifac[i]%mod);
}