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<algorithm>
using namespace std;

typedef long long LL;

const int N=200005;
const int MOD=924844033;

int n,cnt,last[N],size[N],a[N*4],b[N*4],rev[N*4],L,jc[N],ny[N];
struct edge{int to,next;}e[N*2];

int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

void addedge(int u,int v)
{
	e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;
	e[++cnt].to=u;e[cnt].next=last[v];last[v]=cnt;
}

void dfs(int x,int fa)
{
	size[x]=1;a[n]++;
	for (int i=last[x];i;i=e[i].next)
	{
		if (e[i].to==fa) continue;
		dfs(e[i].to,x);
		size[x]+=size[e[i].to];
		a[size[e[i].to]]--;
	}
	if (n-size[x]) a[n-size[x]]--;
}

int C(int n,int m)
{
	return (LL)jc[n]*ny[m]%MOD*ny[n-m]%MOD;
}

int ksm(int x,int y)
{
	int ans=1;
	while (y)
	{
		if (y&1) ans=(LL)ans*x%MOD;
		x=(LL)x*x%MOD;y>>=1;
	}
	return ans;
}

void NTT(int *a,int f)
{
	for (int i=0;i<L;i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
	for (int i=1;i<L;i<<=1)
	{
		int wn=ksm(5,f==1?(MOD-1)/i/2:MOD-1-(MOD-1)/i/2);
		for (int j=0;j<L;j+=(i<<1))
		{
			int w=1;
			for (int k=0;k<i;k++)
			{
				int u=a[j+k],v=(LL)a[j+k+i]*w%MOD;
				a[j+k]=u+v;a[j+k]-=a[j+k]>=MOD?MOD:0;
				a[j+k+i]=u+MOD-v;a[j+k+i]-=a[j+k+i]>=MOD?MOD:0;
				w=(LL)w*wn%MOD;
			}
		}
	}
	int ny=ksm(L,MOD-2);
	if (f==-1) for (int i=0;i<L;i++) a[i]=(LL)a[i]*ny%MOD;
}

int main()
{
	n=read();
	for (int i=1;i<n;i++)
	{
		int x=read(),y=read();
		addedge(x,y);
	}
	jc[0]=jc[1]=ny[0]=ny[1]=1;
	for (int i=2;i<=n;i++) jc[i]=(LL)jc[i-1]*i%MOD,ny[i]=(LL)(MOD-MOD/i)*ny[MOD%i]%MOD;
	for (int i=2;i<=n;i++) ny[i]=(LL)ny[i-1]*ny[i]%MOD;
	dfs(1,0);
	for (int i=0;i<=n;i++) b[i]=ny[i],a[i]+=a[i]<0?MOD:0,a[i]=(LL)a[i]*jc[i]%MOD;
	reverse(b,b+n+1);
	int lg=0;
	for (L=1;L<=n*2;L<<=1,lg++);
	for (int i=0;i<L;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(lg-1));
	NTT(a,1);NTT(b,1);
	for (int i=0;i<L;i++) a[i]=(LL)a[i]*b[i]%MOD;
	NTT(a,-1);
	for (int i=1;i<=n;i++) printf("%d\n",(LL)ny[i]*a[n+i]%MOD);
	return 0;
}