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 K 3000008
#define M 524288
#define N 200001
#define p 924844033
#define ll long long
inline int mod(int x){return x>=p?x-p:x;}
inline int mul(int x,int y){return(ll)x*y%p;}
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;
}
char G[K],*H=G;
inline int read()
{
	int x=0;
	while(*H<48||57<*H)H++;
	while(47<*H&&*H<58)x=(x<<1)+(x<<3)+(*H++^48);
	return x; 
}
int S[10],T;
inline void print(int x)
{
	int y;
	do y=x/10,S[++T]=x-y*10,x=y;while(x);
	while(T)*H++=S[T--]^48;
}
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;
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 I,J,v;
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 i=1;i<m;i<<=1)for(I=0,J=i,u>>=1;J<m;I+=i,J+=i)for(v=m&u;J&i;v+=u)x=o[I],y=mul(w[v],o[J]),o[I++]=mod(x+y),o[J++]=mod(p+x-y);
}
int main()
{
	fread(G,1,K,stdin),s[n=read()]=1;
	for(int i=1,u,v;i<n;i++)s[i]=1,b[++t]=a[u=read()],c[a[u]=t]=v=read(),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]=mul(w[i-1],w[1]);
	for(int i=2;i<=n;i++)d[i]=mul(d[i-1],i),e[i]=mul(p-p/i,e[p%i]);
	for(int i=2;i<=n;i++)e[i]=mul(e[i],e[i-1]);
	for(int i=0;i<=n;i++)l[i]=mul(l[i],d[i]),r[i]=e[n-i];
	NTT(l,m),NTT(r,m),H=G;
	for(int i=0;i<m;i++)l[i]=mul(l[i],r[i]);
	NTT(l,-m),t=calc(m,p-2);
	for(int i=1;i<=n;i++)print(mod(p+mul(mul(mul(d[n],e[n-i]),e[i]),n)-mul(mul(l[n+i],t),e[i]))),*H++='\n';
	fwrite(G,1,H-G,stdout);
	return 0;
}