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;
#define N 210000
#define mod 924844033
#define ll long long
#define g 5
int n,tot,len;
int head[N],nex[N<<1],to[N<<1];
int size[N],jc[N],njc[N],ans[N];
int a[N<<2],b[N<<2],c[N<<2];
int qpow(int x,int y)
{
	int ret=1;
	while(y)
	{
		if(y&1)ret=(ll)ret*x%mod;
		x=(ll)x*x%mod;y>>=1;
	}
	return ret;
}
void init()
{
	jc[0]=njc[0]=1;
	for(int i=1;i<=n;i++)jc[i]=(ll)jc[i-1]*i%mod;
	njc[n]=qpow(jc[n],mod-2);
	for(int i=n-1;i>=1;i--)njc[i]=(ll)njc[i+1]*(i+1)%mod;
	for(len=1;len<(n+1)*2;len<<=1);
}
void add(int x,int y)
{
	tot++;
	nex[tot]=head[x];head[x]=tot;
	to[tot]=y;
}
void dfs(int x,int y)
{
	size[x]=1;
	for(int i=head[x];i;i=nex[i])
		if(to[i]!=y)
		{
			dfs(to[i],x);
			size[x]+=size[to[i]];
		}
}
int C(int x,int y){return (ll)jc[x]*njc[y]%mod*njc[x-y]%mod;}
void NTT(int *a,int len,int type)
{
	for(int i=0,t=0;i<len;i++)
	{
		if(i<t)swap(a[i],a[t]);
		for(int j=len>>1;(t^=j)<j;j>>=1);
	}
	for(int i=2;i<=len;i<<=1)
	{
		int wn=qpow(g,(mod-1)/i);
		for(int j=0;j<len;j+=i)
		{
			int w=1,t;
			for(int k=0;k<i>>1;k++,w=(ll)w*wn%mod)
			{
				t=(ll)w*a[j+k+(i>>1)]%mod;
				a[j+k+(i>>1)]=(a[j+k]-t+mod)%mod;
				a[j+k]=(a[j+k]+t)%mod;
			}
		}
	}
	if(type==-1)
	{
		for(int i=1;i<len>>1;i++)swap(a[i],a[len-i]);
		int t=qpow(len,mod-2);
		for(int i=0;i<len;i++)a[i]=(ll)a[i]*t%mod;
	}
}
int main()
{
	//freopen("tt.in","r",stdin);
	scanf("%d",&n);
	init();
	for(int i=1,x,y;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		add(x,y);add(y,x);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++)
		ans[i]=(ll)n*C(n,i)%mod;
	for(int i=2;i<=n;i++)(a[size[i]]+=jc[size[i]])%=mod;
	for(int i=0;i<=n;i++)b[i]=njc[n-i];
	NTT(a,len,1);NTT(b,len,1);
	for(int i=0;i<len;i++)c[i]=(ll)a[i]*b[i]%mod;
	NTT(c,len,-1);
	for(int i=1;i<=n;i++)
	{
		c[n+i]=(ll)c[n+i]*njc[i]%mod;
		ans[i]=(ans[i]-c[n+i]+mod)%mod;
	}

	memset(a,0,sizeof(a));
	for(int i=2;i<=n;i++)
	{	
		size[i]=n-size[i];
		(a[size[i]]+=jc[size[i]])%=mod;
	}
	NTT(a,len,1);
	for(int i=0;i<len;i++)c[i]=(ll)a[i]*b[i]%mod;
	NTT(c,len,-1);
	for(int i=1;i<=n;i++)
	{
		c[n+i]=(ll)c[n+i]*njc[i]%mod;
		ans[i]=(ans[i]-c[n+i]+mod)%mod;
	}
	for(int i=1;i<=n;i++)
		printf("%d\n",ans[i]);
	return 0;
}
