Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
const int mod=924844033,maxn=200010;
int hea[maxn],nxt[maxn<<1],to[maxn<<1],a[maxn<<2],b[maxn<<2],n,tot;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
	for(;ch>='0'&&ch<='9';x=(x<<1)+(x<<3)+ch-'0',ch=getchar());
	return x*f;
}
void add(int x,int y)
{
	nxt[++tot]=hea[x];
	hea[x]=tot;
	to[tot]=y;
}
int powmod(int a,int b)
{
	int ans=1;
	while(b)
	{
		if(b&1)ans=1ll*ans*a%mod;
		a=1ll*a*a%mod;
		b>>=1;
	}
	return ans;
}
int dfs(int x,int fa)
{
	int s=1;
	for(int i=hea[x];i;i=nxt[i])if(to[i]!=fa)
	{
		int t=dfs(to[i],x);
		a[t]++;
		if(a[t]>=mod)a[t]-=mod;
		s+=t;
	}
	if(s<n)a[n-s]++;
	if(a[n-s]>=mod)a[n-s]-=mod;
	return s;
}
void NTT(int n,int *x,int f)
{
	for(int i=0,j=0;i<n;i++)
	{
		if(j<i)swap(x[i],x[j]);
		for(int l=n>>1;(j^=l)<l;l>>=1);
	}
	for(int i=2;i<=n;i<<=1)
	{
		int m=i>>1,wn=powmod(5,(mod-1)/i);
		if(f==-1)wn=powmod(wn,mod-2);
		for(int j=0;j<n;j+=i)
			for(int k=0,w=1;k<m;k++)
			{
				int t=1ll*w*x[j+k+m]%mod;
				x[j+k+m]=(x[j+k]-t+mod)%mod;
				x[j+k]=(x[j+k]+t)%mod;
				w=1ll*w*wn%mod;
			}
	}
	int t=powmod(n,mod-2);
	if(f==-1)
		for(int i=0;i<n;i++)
			x[i]=1ll*x[i]*t%mod;
}
int main()
{
	n=read();
	int x,y,s=1;
	for(int i=1;i<n;i++)
	{
		x=read(),y=read();
		add(x,y);
		add(y,x);
	}
	dfs(1,0);
	b[0]=1;
	for(int i=1;i<=n;i++)
	{
		s=1ll*s*i%mod;
		a[i]=1ll*a[i]*s%mod;
		b[i]=powmod(s,mod-2);
	}
	for(int i=0;i<=n>>1;i++)
		swap(a[i],a[n-i]);
	int n1=1,t=n<<1;
	while(n1<=t)n1<<=1;
	NTT(n1,a,1),NTT(n1,b,1);
	for(int i=0;i<n1;i++)
		a[i]=1ll*a[i]*b[i]%mod;
	NTT(n1,a,-1);
	int s1=1,s2=1;
	for(int i=1;i<=n;i++)
	{
		s1=1ll*s1*i%mod;
		s2=1ll*s2*(n-i+1)%mod;
		int t=powmod(s1,mod-2);
		printf("%lld\n",(1ll*n*s2%mod*t%mod-1ll*t*a[n-i]%mod+mod)%mod);
	}
	return 0;
}