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<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;++i)
#define fd(i,a,b) for(int i=a;i>=b;--i)
#define efo(i,v,u) for(int i=BB[v],u=B[BB[v]][1];i;u=B[i=B[i][0]][1])
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
using namespace std;
typedef long long ll;typedef double db;
char ch;
int read(){int n=0,p=1;for(ch=getchar();ch<'0'||'9'<ch;ch=getchar())if(ch=='-')p=-1;for(;'0'<=ch && ch<='9';ch=getchar())n=n*10+ch-'0';return n*p;}
const int N=8e5+5,mo=924844033;
void inc(ll &x,ll y){x=(x+y)%mo;}
ll qmi(ll x,ll n)
{
	ll t=1;
	for(x%=mo;n;n>>=1,x=x*x%mo) if(n&1) t=t*x%mo;
	return t;
}
int n,B0,BB[N],B[N][2];
void link(int u,int v){B[++B0][1]=v,B[B0][0]=BB[u],BB[u]=B0;}
ll ans[N],fac[N],iv[N],f[N],g[N];
int size[N];
void dfs(int v,int fr=0)
{
	size[v]=1;
	efo(i,v,u) if(u!=fr)
	{
		dfs(u,v);
		size[v]+=size[u];
	}
	if(size[v]!=n) ++f[size[v]],++f[n-size[v]];
}
int len,lg,bitrev[N];
ll invlen,W[N];
void dft(ll *a,int sig=1)
{
	fo(i,0,len-1) if(i<bitrev[i]) swap(a[i],a[bitrev[i]]);
	for(int i=2,gap=len>>1;i<=len;i<<=1,gap>>=1)	
		for(int j=0;j<len;j+=i)
			for(ll k=0,*l=a+j,*r=a+j+(i>>1),*w=sig>0?W:W+len;k<(i>>1);++k,++l,++r,w=sig>0?w+gap:w-gap)
			{
				ll u=*l,v=*r * *w%mo;
				*l=(u+v)%mo,*r=(u-v+mo)%mo;
			}
	if(sig==-1) fo(i,0,len-1) a[i]=a[i]*invlen%mo;
}
void ntt(ll *c,ll *a,ll *b)
{
	for(len=1,lg=0;len<=n+n;len<<=1,++lg);
	invlen=qmi(len,mo-2);
	W[0]=1,W[1]=qmi(5,(mo-1)/len);
	fo(i,2,len) W[i]=W[i-1]*W[1]%mo;
	fo(i,0,len-1) bitrev[i]=(bitrev[i>>1]>>1)|((i&1)<<(lg-1));
	dft(a);
	dft(b);
	fo(i,0,len-1) c[i]=a[i]*b[i];
	dft(c,-1);
}
int main()
{
	int x,y;
	n=read();
	fac[0]=iv[0]=1;
	fo(i,1,n) fac[i]=fac[i-1]*i%mo;
	iv[n]=qmi(fac[n],mo-2);
	fd(i,n-1,1) iv[i]=iv[i+1]*(i+1)%mo;
	fo(i,1,n-1) x=read(),y=read(),link(x,y),link(y,x);
	dfs(1);
	fo(i,0,n) f[i]=f[i]*fac[i]%mo,g[i]=iv[i];
	reverse(f+1,f+n+1);
	ntt(ans,f,g);
	reverse(ans+1,ans+n+1);
	fo(i,1,n) ans[i]=ans[i]*iv[i]%mo;
	fo(i,1,n) ans[i]=(n*fac[n]%mo*iv[i]%mo*iv[n-i]%mo-ans[i]+mo)%mo;
	fo(i,1,n) printf("%lld\n",ans[i]);
	return 0;
}