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<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
const ll MOD=924844033;
struct Edge
{
	int to;
	int next;
}e[1000005];
int n,m,edgenum,head[200005],pa[200005],size[200005];
int limit=1,l,r[1000005];
ll A[1000005],B[1000005],fact[1000005],inv[1000005];
void add(int u,int v)
{
	e[++edgenum].to=v;
	e[edgenum].next=head[u];
	head[u]=edgenum;
}
ll quick_pow(ll x,ll a)
{
	ll ans=1;
	while(a)
	{
		if(a&1)ans=ans*x%MOD;
		x=x*x%MOD;
		a>>=1;
	}
	return ans;
}
void dfs(int node)
{
	size[node]=1;
	for(int hd=head[node];hd;hd=e[hd].next)
	{
		int to=e[hd].to;
		if(to==pa[node])continue;
		pa[to]=node;
		dfs(to);
		A[n]++;
		A[size[to]]--;
		A[n-size[to]]--;
		size[node]+=size[to];
	}
}
void NTT(ll*A,int type)
{
	for(int i=0;i<limit;i++)
	  if(i<r[i])swap(A[i],A[r[i]]);
	for(int mid=1;mid<limit;mid<<=1)
	{
		ll Wn=quick_pow(5,(MOD-1)/(mid<<1));
		for(int R=mid<<1,j=0;j<limit;j+=R)
		{
			ll w=1;
			for(int k=0;k<mid;k++,w=w*Wn%MOD)
			{
				ll x=A[j+k],y=A[j+mid+k]*w%MOD;
				A[j+k]=x+y;
				A[j+mid+k]=x-y;
				if(A[j+k]>MOD)A[j+k]-=MOD;
				if(A[j+mid+k]<0)A[j+mid+k]+=MOD;
			}
		}
	}
	if(type==-1)
	{
		ll inv=quick_pow(limit,MOD-2);
		for(int i=0;i<limit;i++)A[i]=A[i]*inv%MOD;
		for(int i=1;i<(limit>>1);i++)swap(A[i],A[limit-i]);
	}
}
ll C(ll n,ll m)
{
	return fact[n]*inv[m]%MOD*inv[n-m]%MOD;
}
int main()
{
	scanf("%d",&n);
	fact[0]=1;
	for(int i=1;i<=n;i++)fact[i]=fact[i-1]*i%MOD;
	inv[n]=quick_pow(fact[n],MOD-2);
	for(int i=n-1;i>=0;i--)inv[i]=inv[i+1]*(i+1)%MOD;
	for(int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
		add(v,u);
	}
	dfs(1);
	for(int i=1;i<=n;i++)
	{
		if(A[i]<0)A[i]+=MOD;
		A[i]=A[i]*fact[i]%MOD;
	}
	for(int i=0;i<=n;i++)
	  B[n-i]=inv[i];
	while(limit<=2*n)limit<<=1,l++;
	for(int i=0;i<limit;i++)
	  r[i]=(r[i>>1]>>1)|((i&1)<<l-1);
	NTT(A,1);
	NTT(B,1);
	for(int i=0;i<limit;i++)A[i]=A[i]*B[i]%MOD;
	NTT(A,-1);
	for(int i=1;i<=n;i++)A[i]=(A[n+i]*inv[i]+C(n,i))%MOD;
	for(int i=1;i<=n;i++)
	  printf("%lld\n",A[i]);
	return 0;
}