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 N=8e5+50,P=924844033;

int head[N],etop;
struct Edge
{
	int v,nxt;
}e[N<<1];
void add(int u,int v)
{
	e[++etop].v=v;
	e[etop].nxt=head[u];
	head[u]=etop;
}

int power(int x,int k)
{
	int ret=1;
	for (;k;k>>=1,x=1LL*x*x%P)
		if (k&1) ret=1LL*ret*x%P;
	return ret;
}
namespace NTT
{
	const int G=5;
	
	int n,r[N],a[N],b[N];
	void init(int m)
	{
		for (n=1;n<m;n<<=1) ;
		for (int i=0;i<n;i++)
			r[i]=(r[i>>1]>>1) | (i&1 ? n>>1 : 0);
	}
	void NTT(int *f,int sgn)
	{
		for (int i=0;i<n;i++)
			if (r[i]>i) swap(f[r[i]],f[i]);
		for (int len,siz=2;siz<=n;siz<<=1)
		{
			len=siz>>1;
			int w=power(G,(P-1)/siz);
			if (sgn==-1) w=power(w,P-2);
			for (int l=0;l<n;l+=siz)
	        {
	            int buf=1;
	            for (int i=l;i<l+len;i++)
	            {
	                int t=1LL*buf*f[i+len]%P;
	                f[i+len]=(f[i]-t)%P;
	                f[i]=(f[i]+t)%P;
	                buf=1LL*buf*w%P;
	            }
	        }
		}
		if (sgn==-1)
	    {
	        int inv=power(n,P-2);
	        for (int i=0;i<n;i++)
	            f[i]=1LL*f[i]*inv%P;
	    }
	    for (int i=0;i<n;i++)
	    	f[i]=(f[i]+P)%P;
	}
	void mul(int la,int *ta,int lb,int *tb,int *res)
	{
		init(la+lb-1);
		for (int i=0;i<la;i++)
			a[i]=ta[i];
		for (int i=0;i<lb;i++)
			b[i]=tb[i];
		NTT(a,1); NTT(b,1);
		for (int i=0;i<n;i++)
			a[i]=1LL*a[i]*b[i]%P;
		NTT(a,-1);
		for (int i=0;i<la+lb-1;i++)
			res[i]=a[i];
	}
}

int n,fac[N],ifac[N],siz[N],a[N];
int g[N],h[N],ans[N];
void dfs(int u,int fa)
{
	siz[u]=1;
	for (int v,i=head[u];i;i=e[i].nxt)
	{
		v=e[i].v;
		if (v==fa) continue;
		dfs(v,u);
		siz[u]+=siz[v];
		--a[siz[v]];
	}
	a[n]++;
	a[n-siz[u]]--;
}
void init()
{
	fac[0]=1;
	for (int i=1;i<=n;i++)
		fac[i]=1LL*fac[i-1]*i%P;
	ifac[n]=power(fac[n],P-2);
	for (int i=n-1;i>=0;i--)
		ifac[i]=1LL*ifac[i+1]*(i+1)%P;
}
int main()
{
	scanf("%d",&n);
	init();
	for (int u,v,i=1;i<n;i++)
	{
		scanf("%d%d",&u,&v);
		add(u,v); add(v,u);
	}
	dfs(1,0);
	for (int i=0;i<n;i++)
	{
		g[i]=ifac[i];
		h[i]=1LL*fac[n-i]*a[n-i]%P;
	}
	NTT::mul(n,g,n,h,ans);
	for (int i=1;i<=n;i++)
		printf("%d\n",1LL*ans[n-i]*ifac[i]%P);
	return 0;
}