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>
#define M 524288
#define N 200001
#define p 924844033
inline int calc(int x,int y)
{
	int z=1;
	for(;y;x=1ll*x*x%p,y>>=1)if(y&1)z=1ll*z*x%p;
	return z;
}
int a[N],b[N<<1],c[N<<1],d[N],e[N],f[M],l[M],m,n,r[M],s[N],t,w[M|1],x,y;
inline void NTT(int*o,int u)
{
	for(int i=0;i<m;i++)if(f[i]>i)t=o[f[i]],o[f[i]]=o[i],o[i]=t;
	for(int j=1,v=m&u;j<m;j<<=1,u>>=1)for(int i=0;i<m;i++)x=o[i],y=1ll*w[v]*o[i+j]%p,o[i]=(x+y)%p,o[i+j]=(p+x-y)%p,(i&j-1)==j-1?i+=j,v=m&u:v+=u;
}
void dfs(int u,int v)
{
	for(int i=a[u];i;i=b[i])if(c[i]!=v)dfs(c[i],u),s[u]+=s[c[i]],l[n-s[c[i]]]++,l[s[c[i]]]++;
}
int main()
{
	scanf("%d",&n),s[n]=1;
	for(int i=1,u,v;i<n;i++)s[i]=1,scanf("%d%d",&u,&v),b[++t]=a[u],c[a[u]=t]=v,b[++t]=a[v],c[a[v]=t]=u;
	dfs(m=1,0),t=0;
	while(m<=n<<1)m<<=1,t++;w[*d=d[1]=*e=e[1]=*w=w[m]=1]=calc(5,p>>t--),f[1]=1<<t;
	for(int i=2;i<m;i++)f[i]=f[i>>1]>>1|(i&1)<<t,w[i]=1ll*w[i-1]*w[1]%p;
	for(int i=2;i<=n;i++)d[i]=1ll*d[i-1]*i%p,e[i]=1ll*(p-p/i)*e[p%i]%p;
	for(int i=2;i<=n;i++)e[i]=1ll*e[i-1]*e[i]%p;
	for(int i=0;i<=n;i++)l[i]=1ll*l[i]*d[i]%p,r[i]=e[n-i];
	NTT(l,m>>1),NTT(r,m>>1);
	for(int i=0;i<m;i++)l[i]=1ll*l[i]*r[i]%p;
	NTT(l,-m>>1),t=calc(m,p-2);
	for(int i=1;i<=n;i++)printf("%lld\n",(1ll*d[n]*e[n-i]%p*e[i]%p*n%p+p-1ll*l[n+i]*t%p*e[i]%p)%p);
	return 0;
}