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<cstdio>
#include<algorithm> 
#include<cstring>
#define LL long long
using namespace std;
const int N=2e5+5;
const int M=524288+5;
const int mod=924844033; 
int n,nn,cnt,u,v,ans,first[N],fac[N],inv[N];
int num[N],sz[N],a[M],b[M];
struct edge{int to,next;}e[N*2];
int read()
{
	int x=0,f=1;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
	return x*f;
}
void ins(int u,int v){e[++cnt]=(edge){v,first[u]};first[u]=cnt;}
void dfs(int x,int fa)
{
	sz[x]=1;
	for(int i=first[x];i;i=e[i].next)
	{
		int to=e[i].to;
		if(to==fa)continue;
		dfs(to,x);
		sz[x]+=sz[to];
	}
	if(fa!=-1)a[sz[x]]++,a[n-sz[x]]++;
}
int power(int a,int b)
{
	int ans=1;
	while(b)
	{
		if(b&1)ans=1ll*ans*a%mod;
		a=1ll*a*a%mod;b>>=1;
	}
	return ans;
}
void ntt(int *a,int n,int f)
{
	int k=0;while((1<<k)<n)k++;
	for(int i=0;i<n;i++)
	{
		int t=0;
		for(int j=0;j<k;j++)
			if(i&(1<<j))t|=(1<<(k-j-1));
		if(i<t)swap(a[i],a[t]);
	}
	for(int l=2;l<=n;l<<=1)
	{
		int m=l>>1,nw=power(5,(mod-1)/l);
		if(f==-1)nw=power(nw,mod-2);
		for(int *p=a;p!=a+n;p+=l)
		{
			int w=1;
			for(int i=0;i<m;i++)
			{
				int t=1ll*p[m+i]*w%mod;
				p[m+i]=(p[i]-t+mod)%mod;
				p[i]=(p[i]+t)%mod;
				w=1ll*w*nw%mod;
			}
		}
	}
	if(f==-1)
	{
		int inv=power(n,mod-2);
		for(int i=0;i<n;i++)a[i]=1ll*a[i]*inv%mod;
	}
}
int main()
{
	n=read();
	for(int i=1;i<n;i++)
	{
		u=read();v=read();
		ins(u,v);ins(v,u);
	}
	dfs(1,-1);
	fac[0]=1;
	for(int i=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mod;
	inv[n]=power(fac[n],mod-2);
	for(int i=n;i>=1;i--)inv[i-1]=1ll*inv[i]*i%mod;
	for(int i=1;i<=n;i++)a[i]=1ll*a[i]*fac[i]%mod;
	for(int i=0;i<=n;i++)b[n-i]=inv[i];
	nn=1;while(nn<n+n+1)nn<<=1;
	ntt(a,nn,1);ntt(b,nn,1);
	for(int i=0;i<nn;i++)a[i]=1ll*a[i]*b[i]%mod;
	ntt(a,nn,-1);
	for(int i=1;i<=n;i++)
	{
		ans=1ll*fac[n]*inv[i]%mod*inv[n-i]%mod*n%mod;
		printf("%lld\n",(ans-1ll*a[n+i]*inv[i]%mod+mod)%mod);
	}
	return 0;
}