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>
using namespace std;
const int N=200005,md=924844033;
int n,i,j,k,x,y,f[N],head[N],adj[N*2],nxt[N*2],size[N],ru[N],lmt,len,g[N*4],h[N*4],ome[N*4],rome[N*4],rt[N*4],inv[N*4],jc[N*4],rjc[N*4];
void dfs(int x,int dad)
{
	size[x]=1;
	for(int y=head[x];y;y=nxt[y])
		if(adj[y]!=dad)
		{
			dfs(adj[y],x);
			size[x]+=size[adj[y]];
			ru[size[adj[y]]]++;
		}
	ru[n-size[x]]++;
}
int pw(int a,int b)
{
	int rtn=1;
	while(b)
	{
		if(b&1)
			rtn=1ll*rtn*a%md;
		a=1ll*a*a%md;
		b>>=1;
	}
	return rtn;
}
void ntt(int a[N*4],bool t)
{
	for(int i=0;i<len;++i)
		if(i<rt[i])
			swap(a[i],a[rt[i]]);
	for(int l=2;l<=len;l<<=1)
	{
		int m=l>>1;
		for(int i=0;i<len;i+=l)
			for(int j=0;j<m;++j)
			{
				int w=1ll*(t?ome[len/l*j]:rome[len/l*j])*a[i+j+m]%md;
				a[i+j+m]=(a[i+j]-w+md)%md;
				a[i+j]=(a[i+j]+w)%md;
			}
	}
	if(!t)
		for(int i=0;i<len;++i)
			a[i]=1ll*a[i]*inv[len]%md;
}
int C(int n,int m)
{
	if(m>n)
		return 0;
	return 1ll*jc[n]*rjc[m]%md*rjc[n-m]%md;
}
int main()
{
	scanf("%d",&n);
	inv[1]=1;
	for(i=2;i<=n*4;++i)
		inv[i]=1ll*inv[md%i]*(md-md/i)%md;
	jc[0]=rjc[0]=1;
	for(i=1;i<=n*4;++i)
		jc[i]=1ll*jc[i-1]*i%md,rjc[i]=1ll*rjc[i-1]*inv[i]%md;
	for(i=1;i<n;++i)
	{
		scanf("%d%d",&j,&k);
		adj[i*2-1]=k;
		nxt[i*2-1]=head[j];
		head[j]=i*2-1;
		adj[i*2]=j;
		nxt[i*2]=head[k];
		head[k]=i*2;
	}
	dfs(1,-1);
	lmt=0;
	while((1<<lmt)<=2*n)
		++lmt;
	len=1<<lmt;
	for(i=0,x=pw(5,(md-1)/len),y=pw(x,md-2);i<len;++i)
	{
		if(i==0)
			ome[i]=rome[i]=1;
		else
			ome[i]=1ll*ome[i-1]*x%md,rome[i]=1ll*rome[i-1]*y%md;
		rt[i]=0;
		for(j=0;j<lmt;++j)
			if((1<<j)&i)
				rt[i]|=1<<(lmt-j-1);
	}
	for(i=1;i<=n;++i)
		g[i]=1ll*jc[i]*ru[i]%md;
	for(i=0;i<=n;++i)
		h[i]=rjc[n-i];
	ntt(g,true);
	ntt(h,true);
	for(i=0;i<len;++i)
		g[i]=1ll*g[i]*h[i]%md;
	ntt(g,false);
	for(i=1;i<=n;++i)
		f[i]=(1ll*C(n,i)*n%md-1ll*g[i+n]*rjc[i]%md+md)%md;
	for(i=1;i<=n;++i)
		printf("%d\n",f[i]);
	return 0;
}
