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<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iostream>
#define mmod 924844033
#define N 210000
#define M 550000
#define grt 5
#define LL long long
using namespace std;
LL jc[M],ny[M];
struct node{LL y,nex;}a[2*N];
struct NTT{
	LL a[M],b[M],c[M],bitrev[M],m,g[M],ng[M];
	LL qmod(LL a,LL b)
	{
		LL t=1;
		while(b)
		{
			if(b%2) t=1ll*t*a%mmod;
			b/=2;
			a=1ll*a*a%mmod;
		}
		return t;
	}
	void ntt_pre(LL n)
	{
		LL len=0;
		while((1<<len)<n) len++;
		len--;
		for(LL i=0;i<n;i++)
		{
			LL k=0;
			for(LL j=0;j<=len;j++) if(i&(1<<j)) k|=1<<(len-j);
			bitrev[i]=k;
		}
	}
	void ntt(LL *x,LL n,LL o)
	{
		for(LL i=0;i<n;i++) if(i>bitrev[i]) swap(x[i],x[bitrev[i]]);
		for(LL s=2;s<=n;s*=2)
		{
			if(o==1 && g[s]==0) g[s]=qmod(grt,(mmod-1)/s);
			if(o==-1 && ng[s]==0) ng[s]=qmod(grt,mmod-1-(mmod-1)/s);
			LL wn;
			if(o==1) wn=g[s];
			else wn=ng[s];
			for(LL i=0;i<n;i+=s)
			{
				LL w=1;
				for(LL j=0;j<s/2;j++)
				{
					LL t1=x[i+j],t2=1ll*w*x[i+j+s/2]%mmod;
					x[i+j]=(t1+t2)%mmod;
					x[i+j+s/2]=(t1-t2)%mmod;
					w=1ll*w*wn%mmod;
				}
			}
		}
	}
	void work()
	{
		m*=2;
		LL n=1;
		while(n-1<m) n*=2;
		ntt_pre(n);
		ntt(a,n,1);ntt(b,n,1);
		for(LL i=0;i<n;i++) c[i]=1ll*a[i]*b[i]%mmod;
		ntt(c,n,-1);
		LL ny=qmod(n,mmod-2);
		for(LL i=0;i<n;i++) c[i]=1ll*c[i]*ny%mmod;
		LL oo=1;
	}
}g;
LL fir[N],len,n,siz[N],cnt[N],ans[N];
void ins(LL x,LL y)
{
	a[++len].y=y;a[len].nex=fir[x];fir[x]=len;
}
void dfs(LL x,LL fa)
{
	siz[x]=1;
	for(LL k=fir[x];k;k=a[k].nex)
	{
		LL y=a[k].y;
		if(y==fa) continue;
		dfs(y,x);
		siz[x]+=siz[y];
		cnt[siz[y]]++;
	}
	cnt[n-siz[x]]++;
}
LL C(LL n,LL m)
{
	return 1ll*jc[n]*ny[m]%mmod*ny[n-m]%mmod;
}
int main()
{
	jc[0]=jc[1]=ny[0]=ny[1]=1;
	for(LL i=2;i<M;i++) jc[i]=1ll*jc[i-1]*i%mmod,ny[i]=1ll*(-mmod/i)*ny[mmod%i]%mmod;
	for(LL i=2;i<M;i++) ny[i]=1ll*ny[i-1]*ny[i]%mmod;
	scanf("%lld",&n);
	for(LL i=1;i<n;i++)
	{
		LL x,y;scanf("%lld%lld",&x,&y);
		ins(x,y);ins(y,x);
	}
	dfs(1,0);
	for(LL i=1;i<=n;i++) 
		g.a[i]=cnt[i]*jc[i]%mmod,g.b[i]=ny[n-i];
	g.m=n;
	g.work();
	for(LL i=1;i<=n;i++) ans[i]=1ll*n*C(n,i)%mmod;
	for(LL i=1;i<=n;i++) ans[i]=(ans[i]-(1ll*ny[i]*g.c[n+i]%mmod))%mmod;
	for(LL i=1;i<=n;i++) printf("%lld\n",(ans[i]+mmod)%mmod);
	return 0;
}