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 <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

typedef long long  LL;
const int N=200010,mo=924844033,g=5;
struct edge{int s,t,n;}e[2*N];
int n,h[N],a[N],siz[N],fac[N],ifac[N],A[1<<19],B[1<<19],W[1<<19];

int qpow(int a,int b)
{
	int x=a;  a=1;
	while (b)
		{
			if (b&1)  a=1LL*a*x%mo;
			x=1LL*x*x%mo,b>>=1;
		}
	return a;
}

void NTT(int *a,int n,int op)
{
	for (int i=1,j,p,t; i<n; i++)
		{
			for (j=0,p=i,t=1; t<n; j<<=1,j|=p&1,p>>=1,t<<=1);
			if (i<j)  swap(a[i],a[j]);
		}
	int *w=(op>0?W:W+n/2),u,v;
	for (int len=2; len<=n; len<<=1)
		for (int i=0; i<n; i+=len)
			for (int j=0; j<(len>>1); j++)
				{
					u=a[i+j],v=1LL*a[i+j+(len>>1)]*w[n/len*j]%mo;
					a[i+j]=(u+v)%mo,a[i+j+(len>>1)]=(u+mo-v)%mo;
				}
	if (op==-1)
		{
			int inv=qpow(n,mo-2);
			for (int i=0; i<n; i++)  a[i]=1LL*a[i]*inv%mo;
		}
}

void FFT()
{
	int N=1;
	while (N<=2*n)  N<<=1;
	W[0]=1,W[1]=qpow(g,(mo-1)/N);
	for (int i=2; i<N; i++)  W[i]=1LL*W[i-1]*W[1]%mo;
	reverse(W+N/2+1,W+N),W[N/2]=1;
	NTT(A,N,1),NTT(B,N,1);
	for (int i=0; i<N; i++)  A[i]=1LL*A[i]*B[i]%mo;
	NTT(A,N,-1);
}

void dfs(int x)
{
	siz[x]=1;
	for (int i=h[x]; i; i=e[i].n)
		if (!siz[e[i].t])  dfs(e[i].t),siz[x]+=siz[e[i].t];
	if (siz[x]!=n)  a[siz[x]]--,a[n-siz[x]]--;
}

void work()
{
	scanf("%d",&n);
	for (int i=1,u,v,tot=0; i<n; i++)
		{
			scanf("%d %d",&u,&v);
			e[++tot]=(edge){u,v,h[u]},h[u]=tot;
			e[++tot]=(edge){v,u,h[v]},h[v]=tot;
		}
	dfs(rand()%n+1),a[n]=n,fac[0]=1;
	for (int i=1; i<=n; i++)  fac[i]=1LL*fac[i-1]*i%mo;
	ifac[n]=qpow(fac[n],mo-2);
	for (int i=n; i; i--)  ifac[i-1]=1LL*ifac[i]*i%mo;
	for (int i=1; i<=n; i++)  A[i]=1LL*(a[i]+mo)*fac[i]%mo;
	for (int i=1; i<=n; i++)  B[i]=ifac[n-i];
	FFT();
	for (int i=1; i<=n; i++)
		printf("%d\n",int(1LL*A[n+i]*ifac[i]%mo));
}

int main()
{
	work();
	return 0;
}
