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<algorithm>
#define N (800000+1000)
#define MOD (924844033)
#define LL long long
using namespace std;

struct Edge{int to,next;}edge[N];
LL n,u,v,fn=1,l,r[N],a[N],b[N];
LL fac[N],inv[N],facinv[N],ans[N];
LL head[N],num_edge,size[N],cnt[N];

void add(int u,int v)
{
	edge[++num_edge].to=v;
	edge[num_edge].next=head[u];
	head[u]=num_edge;
}

LL Qpow(LL a,LL b)
{
	LL ans=1,base=a;
	while (b)
	{
		if (b&1) ans=ans*base%MOD;
		base=base*base%MOD; b>>=1;
	}
	return ans;
}

void NTT(LL n,LL *a,LL opt)
{
	for (int i=0; i<n; ++i)
		if (i<r[i]) swap(a[i],a[r[i]]);
	for (int k=1; k<n; k<<=1)
	{
		LL wn=Qpow(5,(MOD-1)/(k<<1));
		for (int i=0; i<n; i+=(k<<1))
		{
			LL w=1;
			for (int j=0; j<k; ++j, w=w*wn%MOD)
			{
				LL x=a[i+j], y=w*a[i+j+k]%MOD;
				a[i+j]=(x+y)%MOD; a[i+j+k]=(x-y+MOD)%MOD;
			}
		}
	}
	if (opt==-1)
	{
		reverse(a+1,a+n);
		LL g=Qpow(n,MOD-2);
		for (int i=0; i<n; ++i) a[i]=a[i]*g%MOD;
	}
}

void Init()
{
	fac[0]=1; inv[1]=1; facinv[0]=1;
	for (int i=1; i<=n; ++i)
	{
		if (i!=1) inv[i]=(MOD-MOD/i)*inv[MOD%i]%MOD;
		fac[i]=fac[i-1]*i%MOD; facinv[i]=facinv[i-1]*inv[i]%MOD;
	}
}

LL C(LL n,LL m)
{
	if (n<m) return 0;
	return fac[n]*facinv[m]%MOD*facinv[n-m]%MOD;
}

void Dfs(int x,int fa)
{
	size[x]=1;
	for (int i=head[x]; i; i=edge[i].next)
		if (edge[i].to!=fa)
		{
			Dfs(edge[i].to,x); 
			cnt[size[edge[i].to]]++;
			size[x]+=size[edge[i].to];
		}
	cnt[n-size[x]]++;
}

int main()
{
	scanf("%lld",&n);
	for (int i=1; i<=n-1; ++i)
	{
		scanf("%lld%lld",&u,&v);
		add(u,v); add(v,u);
	}
	Init(); Dfs(1,-1);
	while (fn<=n+n) fn<<=1, l++;
	for (int i=0; i<=fn; ++i)
		r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
		
	for (int i=0; i<=n; ++i) a[i]=cnt[i]*fac[i]%MOD;
	for (int i=0; i<=n; ++i) b[i]=facinv[n-i];

	NTT(fn,a,1); NTT(fn,b,1);
	for (int i=0; i<=fn; ++i) a[i]=a[i]*b[i]%MOD;
	NTT(fn,a,-1);
	
	for (int i=1; i<=n; ++i) ans[i]=-facinv[i]*a[n+i]%MOD+MOD;
	for (int i=1; i<=n; ++i) ans[i]=(ans[i]+n*C(n,i))%MOD;
	for (int i=1; i<=n; ++i) printf("%lld\n",ans[i]);
}