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<cstring>
#include<cstdlib>
#include<algorithm>

using namespace std;

inline int read()
{
	int x=0;char ch=getchar();
	while(ch<'0' || '9'<ch)ch=getchar();
	while('0'<=ch && ch<='9')x=x*10+(ch^48),ch=getchar();
	return x;
}

typedef long long ll;
const int N=400009;
const ll md=924844033;

int n,m,l;
int to[N<<1],nxt[N<<1],beg[N],tot;
int rev[N<<1],siz[N];
ll a[N<<1],b[N<<1],fac[N],inv[N];

inline void chk(ll &a){if(a<0)a+=md;}

inline ll qpow(ll a,ll b)
{
	ll ret=1ll;
	while(b)
	{
		if(b&1ll)
			ret=ret*a%md;
		a=a*a%md;
		b>>=1;
	}
	return ret;
}

inline void add(int u,int v)
{
	to[++tot]=v;
	nxt[tot]=beg[u];
	beg[u]=tot;
}

inline void dfs(int u,int fa)
{
	a[n]++;siz[u]=1;
	for(int i=beg[u];i;i=nxt[i])
		if(to[i]!=fa)
		{
			dfs(to[i],u);
			siz[u]+=siz[to[i]];
			chk(--a[siz[to[i]]]);
		}
	chk(--a[n-siz[u]]);
}

inline void NTT(ll *a,int n,bool f)
{
	for(int i=0;i<n;i++)
		if(i<rev[i])
			swap(a[i],a[rev[i]]);
	for(int h=2;h<=n;h<<=1)
	{
		ll w=qpow(5,(md-1)/h);
		if(f)w=qpow(w,md-2);
		for(int j=0;j<n;j+=h)
		{
			ll wn=1;
			for(int k=j;k<j+(h>>1);k++)
			{
				ll x=a[k],y=wn*a[k+(h>>1)]%md;
				a[k]=(x+y)%md;
				a[k+(h>>1)]=(x-y+md)%md;
				wn=wn*w%md;
			}
		}
	}

	if(f)
		for(ll i=0,invn=qpow(n,md-2);i<n;i++)
			a[i]=a[i]*invn%md;
}

inline void init()
{
	fac[0]=1;
	for(ll i=1;i<N;i++)
		fac[i]=fac[i-1]*i%md;
	inv[N-1]=qpow(fac[N-1],md-2);
	for(ll i=N-1;i>=1;i--)
		inv[i-1]=inv[i]*i%md;
}

int main()
{
	if(fopen("vj.in","r"))
	{
		freopen("vj.in","r",stdin);
		freopen("vj.out","w",stdout);
	}

	init();
	n=read();
	for(int i=1,u,v;i<n;i++)
	{
		u=read();v=read();
		add(u,v);add(v,u);
	}
	dfs(1,1);

	for(int i=1;i<=n;i++)
		a[i]=a[i]*fac[i]%md;
	for(int i=1;i<=n;i++)
		b[i]=inv[n-i];

	for(m=1,l=0;m<=(n<<1);m<<=1)l++;
	for(int i=0;i<m;i++)
		rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));

	NTT(a,m,0);NTT(b,m,0);
	for(int i=0;i<m;i++)
		a[i]=a[i]*b[i]%md;
	NTT(a,m,1);

	for(int i=1;i<=n;i++)
		printf("%lld\n",a[n+i]*inv[i]%md);
	return 0;
}
