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<bits/stdc++.h>
using namespace std;
const int N=11e5+10,mod=924844033;
int qp(int x,int k)
{
	int res=1;
	for(;k;k>>=1,x=1ll*x*x%mod)
		if(k&1)
			res=1ll*x*res%mod;
	return res;
}
int n,rev[N],lim,tmp,len;
int head[N],tot;
struct edge
{
	int to,next;
}e[N<<1];
void adde(int a,int b)
{
	e[++tot].to=b; e[tot].next=head[a]; head[a]=tot;
}
int cnt[N],ans[N];
int dfs(int k,int fa)
{
	int sz=1;
	for (int i=head[k];i;i=e[i].next)
		if(e[i].to!=fa)
			sz+=dfs(e[i].to,k);
	if(fa)cnt[sz]++,cnt[n-sz]++;
	return sz;
}
int fac[N],inv[N],num[N];
void init()
{
	lim=N-10;
	fac[0]=inv[0]=1;
	for(int i=1;i<=lim;i++)
		fac[i]=1ll*fac[i-1]*i%mod;
	inv[lim]=qp(fac[lim],mod-2);
	for(int i=lim-1;i;i--)
		inv[i]=1ll*inv[i+1]*(i+1)%mod;
}
void ntt(int f[],int opt)
{
	for(int i=0;i<lim;i++)
		if(i<rev[i])
			swap(f[i],f[rev[i]]);
	for(int i=1;i<lim;i<<=1)
	{
		int tmp=i<<1;
		int w=qp(5,opt?(mod-1)/tmp:mod-1-(mod-1)/tmp);
		for(int j=0;j<lim;j+=tmp)
		{
			int wn=1;
			for(int k=0;k<i;k++,wn=1ll*w*wn%mod)
			{
				int x=1ll*wn*f[i+j+k]%mod;
				f[i+j+k]=(f[j+k]-x+mod)%mod;
				f[j+k]=(f[j+k]+x)%mod;
			}
		}
	}
	if(!opt)
	{
		int inv=qp(lim,mod-2);
		for(int i=0;i<lim;i++)
			f[i]=1ll*f[i]*inv%mod;
	}
}
int main()
{
	scanf("%d",&n);
	int u,v;
	for(int i=1;i<n;i++)
		scanf("%d%d",&u,&v),adde(v,u),adde(u,v);
	init();
	dfs(1,0);
	for(int i=1;i<=n;i++)
		cnt[i]=1ll*cnt[i]*fac[i]%mod;
	for(int i=0;i<=n;i++)
		num[n-i]=inv[i];
	for(len=n+n+1,lim=1;lim<=len;lim<<=1,tmp++);
	for(int i=0;i<lim;i++)
		rev[i]=(rev[i>>1]>>1)|((i&1)<<tmp-1);
	ntt(cnt,1); ntt(num,1);
	for(int i=0;i<lim;i++)
		ans[i]=1ll*cnt[i]*num[i]%mod;
	ntt(ans,0);
	for(int i=1;i<=n;i++)
		printf("%lld\n",(1ll*fac[n]*inv[i]%mod*inv[n-i]%mod*n%mod-1ll*ans[i+n]*inv[i]%mod+mod)%mod);
	return 0;
}
