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<cstring>
namespace Math
{
	typedef long long ll;
	const int N=200010,M=525000,MOD=924844033;
	ll qpow(ll a,ll b){ll c=1;for(;b;b>>=1,a=a*a%MOD)if(b&1)c=c*a%MOD;return c;}
	int dec[M];
	void getdec(int n){dec[0]=0;for(int i=1;i<n;i++)dec[i]=(dec[i>>1]>>1)+((i&1)?(n>>1):0);}
	void ntt(ll *y,int n,int rag)
	{
		for(int i=0;i<n;i++)
			if(dec[i]<i)std::swap(y[i],y[dec[i]]);
		for(int len=2;len<=n;len<<=1)
		{
			int h=len>>1;
			ll delta=qpow(5,(MOD-1)/len),x,*y0,*y1,ny0,ny1;

			if(rag<0)delta=qpow(delta,MOD-2);

			for(int p=0;p<n;p+=len)
			{
				y0=y+p,y1=y0+h,x=1;
				for(int i=0;i<h;i++,x=x*delta%MOD)
				{
					ny0=(y0[i]+x*y1[i])%MOD;
					ny1=(y0[i]-x*y1[i])%MOD;
					y0[i]=ny0;
					y1[i]=ny1;
				}
			}
		}
	}
	ll A[M],B[M];
	int m;
	void multi()
	{
		getdec(m);
		ntt(A,m,1),ntt(B,m,1);
		for(int i=0;i<m;i++)
			A[i]=A[i]*B[i]%MOD;
		ntt(A,m,-1);
		ll tmp=qpow(m,MOD-2);
		for(int i=0;i<m;i++)
			A[i]=(A[i]*tmp%MOD+MOD)%MOD;
	}

	ll fact[N],ifact[N];
	void calc(int *b,int *ans,int n)
	{
		for(m=1;m<=n+n;m<<=1);
		for(int i=fact[0]=1;i<=n;i++)fact[i]=fact[i-1]*i%MOD;
		ifact[n]=qpow(fact[n],MOD-2);
		for(int i=n;i;i--)ifact[i-1]=ifact[i]*i%MOD;

		for(int i=1;i<=n;i++)A[i]=b[i]*fact[i]%MOD;
		for(int i=0;i<=n;i++)B[i]=ifact[n-i];
		multi();

		for(int i=1;i<=n;i++)
			ans[i]=(A[n+i]*ifact[i]%MOD+MOD)%MOD;
	}
}
namespace O_O
{
	typedef long long ll;
	const int N=201000,M=N*2,MOD=924844033;
	int begin[N],next[M],to[M];
	int cnt[N],ans[N];
	int n,e;
	void add(int x,int y,bool k=1)
	{
		to[++e]=y;
		next[e]=begin[x];
		begin[x]=e;
		if(k)add(y,x,0);
	}
	void initialize()
	{
		scanf("%d",&n);
		for(int i=1,u,v;i<n;i++)
			scanf("%d%d",&u,&v),add(u,v);
	}

	int dfs(int p=1,int h=0)
	{
		int sizp=1,x;
		for(int i=begin[p],q;i;i=next[i])
			if((q=to[i])!=h)
			{
				sizp+=(x=dfs(q,p));
				cnt[x]--;
				cnt[n-x]--;
			}
		return sizp;
	}
	void solve()
	{
		initialize();
		dfs();
		cnt[n]=n;
		Math::calc(cnt,ans,n);
		for(int i=1;i<=n;i++)
			printf("%d\n",ans[i]);
	}
}
int main()
{
	O_O::solve();
	return 0;
}
