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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
#define MOD 924844033
#define ll long long
#define MAX 888888
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
struct Line{int v,next;}e[MAX];
int h[MAX],cnt=1;
inline void Add(int u,int v){e[cnt]=(Line){v,h[u]};h[u]=cnt++;}
int n,f[MAX],g[MAX],size[MAX];
int jc[MAX],jv[MAX],inv[MAX];
void dfs(int u,int ff)
{
	size[u]=1;
	for(int i=h[u];i;i=e[i].next)
	{
		int v=e[i].v;if(v==ff)continue;
		dfs(v,u);size[u]+=size[v];
		++f[size[v]];
	}
	++f[n-size[u]];
}
int C(int n,int m){if(m>n)return 0;return 1ll*jc[n]*jv[m]%MOD*jv[n-m]%MOD;}
int N,l,r[MAX],W[MAX];
int fpow(int a,int b)
{
	int s=1;
	while(b){if(b&1)s=1ll*s*a%MOD;a=1ll*a*a%MOD;b>>=1;}
	return s;
}
void NTT(int *P,int opt)
{
	for(int i=1;i<N;++i)if(i>r[i])swap(P[i],P[r[i]]);
	for(int i=1;i<N;i<<=1)
	{
		int w=fpow(5,(MOD-1)/(i<<1));W[0]=1;
		for(int k=1;k<i;++k)W[k]=1ll*W[k-1]*w%MOD;
		for(int p=i<<1,j=0;j<N;j+=p)
			for(int k=0;k<i;++k)
			{
				int X=P[j+k],Y=1ll*P[i+j+k]*W[k]%MOD;
				P[j+k]=(X+Y)%MOD;P[i+j+k]=(X+MOD-Y)%MOD;
			}
	}
	if(opt==-1)
	{
		reverse(&P[1],&P[N]);
		for(int i=0,inv=fpow(N,MOD-2);i<N;++i)P[i]=1ll*P[i]*inv%MOD;
	}
}
int main()
{
	n=read();
	for(int i=1;i<n;++i)
	{
		int u=read(),v=read();
		Add(u,v);Add(v,u);
	}
	dfs(1,0);
	jc[0]=jv[0]=inv[0]=inv[1]=1;
	for(int i=1;i<=n;++i)jc[i]=1ll*jc[i-1]*i%MOD;
	for(int i=2;i<=n;++i)inv[i]=1ll*inv[MOD%i]*(MOD-MOD/i)%MOD;
	for(int i=1;i<=n;++i)jv[i]=1ll*jv[i-1]*inv[i]%MOD;
	for(N=1;N<n+n;N<<=1)++l;
	for(int i=0;i<N;++i)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
	for(int i=0;i<=n;++i)f[i]=1ll*f[i]*jc[i]%MOD;
	for(int i=0;i<=n;++i)g[i]=jv[n-i];
	NTT(f,1);NTT(g,1);
	for(int i=0;i<N;++i)f[i]=1ll*f[i]*g[i]%MOD;
	NTT(f,-1);
	for(int i=1;i<=n;++i)f[i]=1ll*f[i+n]*jv[i]%MOD;
	for(int i=1;i<=n;++i)f[i]=(1ll*n*C(n,i)%MOD+MOD-f[i])%MOD;
	for(int i=1;i<=n;++i)printf("%d\n",f[i]);
	return 0;
}
