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<cctype>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<ctime>
#include<cstdlib>
#include<algorithm>
#define N 1100000
#define L 1000000
#define eps 1e-7
#define inf 1e9+7
#define db double
#define ll long long
#define ldb long double
using namespace std;
inline ll read()
{
	char ch=0;
	ll x=0,flag=1;
	while(!isdigit(ch)){ch=getchar();if(ch=='-')flag=-1;}
	while(isdigit(ch)){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*flag;
}
const ll d=5,mo=924844033;
ll ksm(ll x,ll k)
{
	ll ans=1;
	while(k)
	{
		if(k&1)ans=ans*x%mo;
		k>>=1;
		x=x*x%mo;
	}
	return ans;
}
ll rev[N];
void ntt(ll *f,ll n,ll flag)
{
	for(ll i=0;i<n;i++)
	{
		rev[i]=(rev[i>>1]>>1)+(i&1)*(n>>1);
		if(i<rev[i])swap(f[i],f[rev[i]]);
	}
	for(ll k=2,kk=1;k<=n;k<<=1,kk<<=1)
	{
		ll wn=ksm(d,(mo-1)/k);
		if(flag==-1)wn=ksm(wn,mo-2)%mo;
		for(ll i=0;i<n;i+=k)
		for(ll j=0,w=1;j<kk;j++,w=1ll*w*wn%mo)
		{
			ll t=1ll*w*f[i+j+kk]%mo;
			f[i+j+kk]=(f[i+j]-t+mo)%mo;
			f[i+j]=(f[i+j]+t)%mo; 
		}
	}
	if(flag==-1)
	{
		ll inv=ksm(n,mo-2);
		for(ll i=0;i<n;i++)f[i]=f[i]*inv%mo;
	}
}
ll a[N],b[N];
void mul(ll n)
{
	ll len;
	for(len=1;len<2*n;len<<=1);
	ntt(a,len,+1);ntt(b,len,+1);
	for(ll i=0;i<len;i++)a[i]=a[i]*b[i]%mo;
	ntt(a,len,-1);
}
struct edge{ll to,nxt;}e[N*2];
ll num,head[N];
inline void add(ll x,ll y){e[++num]=(edge){y,head[x]};head[x]=num;}
ll sz[N],cnt[N],fac[N],vac[N];
void dfs(ll x,ll fa)
{
	sz[x]=1;
	for(ll i=head[x];i!=-1;i=e[i].nxt)
	{
		ll to=e[i].to;
		if(to==fa)continue;
		dfs(to,x);
		sz[x]+=sz[to];
	}
}
ll c(ll n,ll m){return n>=m?(fac[n]*vac[m]%mo*vac[n-m]%mo):0;}
int main()
{
	num=-1;memset(head,-1,sizeof(head));
	ll n=read();
	for(ll i=1;i<=n-1;i++){ll x=read(),y=read();add(x,y);add(y,x);}
	dfs(1,1);for(ll i=2;i<=n;i++)cnt[sz[i]]++,cnt[n-sz[i]]++;
	fac[0]=vac[0]=1;for(ll i=1;i<=L;i++)fac[i]=fac[i-1]*i%mo,vac[i]=vac[i-1]*ksm(i,mo-2)%mo;
	for(ll i=0;i<=n;i++)a[i]=1ll*cnt[i]*fac[i]%mo,b[i]=vac[n-i];
	mul(n+1);
	for(ll i=1;i<=n;i++)printf("%lld\n",((n*c(n,i)%mo)-(vac[i]*a[n+i]%mo)+mo)%mo);
	return 0;
}