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<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<iostream>
#include<map>
#include<string>

#define maxn 400005
#define MOD 924844033

using namespace std;

inline int getint()
{
	int num=0,flag=1;char c;
	while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;
	while(c>='0'&&c<='9')num=num*10+c-48,c=getchar();
	return num*flag;
}

int n;
int fir[maxn],nxt[maxn],to[maxn],cnt;
int rev[maxn<<2];
int sz[maxn],a[maxn<<2],b[maxn<<2];
int fac[maxn],inv[maxn];

inline void newnode(int u,int v)
{to[++cnt]=v,nxt[cnt]=fir[u],fir[u]=cnt;}
inline void dfs(int u,int fa)
{
	sz[u]=1;
	for(int i=fir[u];i;i=nxt[i])if(to[i]!=fa)
	{
		dfs(to[i],u),sz[u]+=sz[to[i]];
		a[sz[to[i]]]++;
	}
	a[n-sz[u]]++;
}

inline int ksm(int num,int k)
{
	int ret=1;
	for(;k;k>>=1,num=1ll*num*num%MOD)if(k&1)ret=1ll*ret*num%MOD;
	return ret;
}
inline void getrev(int N)
{for(int i=0;i<N;i++)rev[i]=(rev[i>>1]>>1)|(i&1?N>>1:0);}
inline void NTT(int *a,int opt,int N)
{
	for(int i=0;i<N;i++)if(rev[i]<i)swap(a[i],a[rev[i]]);
	for(int i=1;i<N;i<<=1)
	{
		int wn=ksm(5,(MOD-1)/(i<<1));
		if(!~opt)wn=ksm(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 x=a[j+k],y=1ll*a[i+j+k]*w%MOD;
			a[j+k]=(x+y)%MOD,a[i+j+k]=(x-y+MOD)%MOD;
		}
	}
	if(!~opt)for(int i=0,Inv=ksm(N,MOD-2);i<N;i++)a[i]=1ll*a[i]*Inv%MOD;
}

int main()
{
	n=getint();
	for(int i=1;i<n;i++)
	{
		int u=getint(),v=getint();
		newnode(u,v),newnode(v,u);
	}
	fac[0]=fac[1]=inv[0]=inv[1]=1;
	for(int i=2;i<=n;i++)fac[i]=1ll*fac[i-1]*i%MOD;
	for(int i=2;i<=n;i++)inv[i]=1ll*inv[MOD%i]*(MOD-MOD/i)%MOD;
	for(int i=2;i<=n;i++)inv[i]=1ll*inv[i]*inv[i-1]%MOD;
	dfs(1,0);a[0]=0;
	for(int i=0;i<=n;i++)a[i]=1ll*a[i]*fac[i]%MOD;
	for(int i=0;i<=n;i++)b[i]=inv[n-i];
	int len=1;
	while(len<=2*n)len<<=1;
	getrev(len);
	NTT(a,1,len),NTT(b,1,len);
	for(int i=0;i<len;i++)a[i]=1ll*a[i]*b[i]%MOD;
	NTT(a,-1,len);
	for(int i=1;i<=n;i++)
	{
		int ans=1ll*fac[n]*inv[i]%MOD*inv[n-i]%MOD*n%MOD;
		printf("%lld\n",(ans-1ll*a[i+n]*inv[i]%MOD+MOD)%MOD);
	}
}