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 long long mod=924844033;
long long n,x,y,sub[222222],coef[222222],jc[222222],injc[222222],sz,cnt,rev[888888],s[888888],t[888888];
vector<long long> gr[222222];
long long binpow(long long a,long long t)
{
	long long res=1,p=a;
	for (long long i=t;i;i>>=1)
	{
		if (i&1) res=res*p%mod;
		p=p*p%mod;
	}
	return res;
}
void Init()
{
	jc[0]=injc[0]=1;
	for (long long i=1;i<=n;i++)
	{
		jc[i]=jc[i-1]*i%mod;
		injc[i]=injc[i-1]*binpow(i,mod-2)%mod;
	}
}
void dfs(long long i,long long fa)
{
	sub[i]=1;
	long long cnt=n-1;
	for (long long j=0;j<gr[i].size();j++)
	{
		long long to=gr[i][j];
		if (to==fa) continue;
		dfs(to,i);
		sub[i]+=sub[to];
		coef[sub[to]]--;cnt-=sub[to];
	}
	if (fa) coef[cnt]--;
}
void ntt(long long c[],long long flag)
{
	for (long long i=0;i<sz;i++)
	{
		if (rev[i]>i) swap(c[i],c[rev[i]]);
	}
	for (long long i=2;i<=sz;i<<=1)
	{
		for (long long j=0;j<sz;j+=i)
		{
			long long w,wn;
			w=1;
			if (flag==1) wn=binpow(5ll,(mod-1)/i);
			else wn=binpow(5ll,(mod-1)-(mod-1)/i);
			for (long long k=j;k<j+i/2;k++)
			{
				long long tmp1=c[k];
				long long tmp2=c[k+i/2]*w%mod;
				c[k]=(tmp1+tmp2)%mod;
				c[k+i/2]=(tmp1-tmp2)%mod;
				w=w*wn%mod;
			}
		}
	}
	if (flag==-1)
	{
		for (long long i=0;i<sz;i++)
		{
			c[i]=c[i]*binpow(sz,mod-2)%mod;
		}
	}
}
void doit()
{
	rev[0]=0;
	for (long long i=1;i<sz;i++)
	{
		rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));
	}
	ntt(s,1);
	ntt(t,1);
	for (long long i=0;i<sz;i++) s[i]=s[i]*t[i]%mod;
	ntt(s,-1);
}
int main()
{
	scanf("%lld",&n);
	for (long long i=1;i<n;i++)
	{
		scanf("%lld%lld",&x,&y);
		gr[x].push_back(y);
		gr[y].push_back(x);  
	}
	Init();
	dfs(1,0);
	coef[n]+=n;
	for (long long i=1;i<=n;i++)
	{
		s[i]=coef[i]*jc[i]%mod;
		t[i]=injc[n-i];
	}
	sz=1;
	while(sz<=n+n)
	{
		cnt++;
		sz*=2;
	}
	doit();
	for (long long i=n+1;i<=n+n;i++) printf("%lld\n",((s[i]*injc[i-n]%mod)+mod)%mod);
	return 0;
}