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<cstdio>
#include<algorithm>
#define LL long long
#define mod 924844033
#define g 5
#define maxn 2000000
using namespace std;
LL n,siz[maxn],f[maxn],fact[maxn],inv[maxn],a[maxn],b[maxn],c[maxn],ans[maxn];
struct node { LL v; node *nxt; } edge[maxn],*head[maxn],*ncnt;
void addedge(LL u,LL v)
{
	ncnt++;
	ncnt->v=v,ncnt->nxt=head[u];
	head[u]=ncnt;
}
void dfs(LL u,LL fa)
{
	siz[u]=1;
	for(node *p=head[u];p;p=p->nxt)
	{
		LL v=p->v;
		if(v==fa) continue;
		dfs(v,u); siz[u]+=siz[v];
	}
	f[siz[u]]--,f[n-siz[u]]--;
}
LL powmod(LL a,LL n)
{
	LL res=1;
	while(n)
	{
		if(n&1) res=1ll*a*res%mod;
		a=1ll*a*a%mod;
		n>>=1;
	}
	return res;
}
void ntt(LL a[],LL n,LL mode)
{
	for(LL i=0,j=0;i<n;i++)
	{
		if(i<j) swap(a[i],a[j]);
		LL k=(n>>1);
		while(k&&(k&j)){ j^=k; k>>=1; }
		j^=k;
	}
	for(LL i=1;i<n;i<<=1)
	{
		LL w1=powmod(g,(mod-1)/(i<<1)),w=1;
		if(mode==-1) w1=powmod(w1,mod-2);
		for(LL j=0;j<i;j++,w=1ll*w*w1%mod)
			for(LL l=j,r=l+i;l<n;l+=(i<<1),r=l+i)
			{
				LL temp=1ll*w*a[r]%mod;
				a[r]=(a[l]-temp+mod)%mod;
				a[l]=(a[l]+temp)%mod;
			}
	}
	if(mode==-1)
	{
		LL inv=powmod(n,mod-2);
		for(LL i=0;i<n;i++) a[i]=1ll*a[i]*inv%mod;
	}
}
int main()
{
	ncnt=&edge[0];
	scanf("%lld",&n);
	for(LL i=1;i<n;i++)
	{
		LL u,v;
		scanf("%lld%lld",&u,&v);
		addedge(u,v); addedge(v,u);
	}
	dfs(1,0);
	f[n]=n;
	fact[0]=1;
	for(LL i=1;i<=n*2;i++) fact[i]=1ll*fact[i-1]*i%mod;
	inv[1]=1;
	for(LL i=2;i<=n*2;i++) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
	inv[0]=1;
	for(LL i=1;i<=n*2;i++) inv[i]=1ll*inv[i-1]*inv[i]%mod;
	for(LL i=0;i<=n;i++) a[i]=1ll*f[i]*fact[i]%mod;
	for(LL i=0;i<=n;i++) b[i]=inv[n-i];
    LL len=1;
    for(;len<=n*2;len<<=1);
    ntt(a,len,1); ntt(b,len,1);
	//for(LL i=0;i<=n*2;i++) printf(" - %lld %lld\n",a[i],b[i]);
    for(LL i=0;i<=len;i++) c[i]=1ll*a[i]*b[i]%mod;
    ntt(c,len,-1);
	for(LL i=0;i<=n;i++) ans[i]=1ll*c[i+n]*inv[i]%mod;
	for(LL i=1;i<=n;i++) printf("%lld\n",ans[i]);
}