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<bits/stdc++.h>
using namespace std;
inline int read(){
	register int x=0;
	register char ch=getchar();
	while(ch<'0'||ch>'9')	ch=getchar();
	while(ch>='0'&&ch<='9')	x=x*10+(ch^48),ch=getchar();
	return x;
}
const int N=1e6+5;
const int P=924844033;
long long Inv,g[N],f[N],fac[N],inv[N];
int n,m,now,len,v[N],r[N],si[N],nex[N],head[N];
void addedges(int x,int y){
	nex[++now]=head[x];
	head[x]=now,v[now]=y;
}
long long C(int n,int m){
	return fac[n]*inv[m]%P*inv[n-m]%P;
}
long long fsp(long long x,int y){
	long long ans=1;
	while(y){
		if(y&1)	ans=ans*x%P;
		x=x*x%P,y>>=1;
	}
	return ans;
}
void NTT(long long *d,int f){
	for(int i=1;i<n;i++)
		if(i<r[i])	swap(d[i],d[r[i]]);
	for(int i=1;i<n;i<<=1){
		int wn=fsp(5,(P-1)/(i<<1));
		if(f)	wn=fsp(wn,P-2);
		for(int j=0;j<n;j+=i<<1){
			long long w=1;
			for(int k=0;k<i;k++,w=w*wn%P){
				int x=d[j+k],y=w*d[i+j+k]%P;
				d[j+k]=(x+y)%P,d[i+j+k]=(x-y+P)%P;
			}
		}
	}
	if(f)	for(int i=0;i<n;i++)	d[i]=d[i]*Inv%P;
}
void dfs(int x,int fa){
	for(int i=head[x];i;i=nex[i])
		if(v[i]!=fa)	dfs(v[i],x),si[x]+=si[v[i]];
	si[x]++,++f[n-si[x]];
	for(int i=head[x];i;i=nex[i])
		if(v[i]!=fa) ++f[si[v[i]]];
}
int main(){
	n=read();
	for(int i=1;i<n;i++){
		int x=read(),y=read();
		addedges(x,y),addedges(y,x);
	}	dfs(1,1),inv[0]=fac[0]=1,f[0]=0;
	for(int i=1;i<=n;i++)	fac[i]=fac[i-1]*i%P;
	for(int i=1;i<=n;i++)	inv[i]=fsp(fac[i],P-2);
	for(int i=0;i<=n;i++)	f[i]=f[i]*fac[i]%P;
	for(int i=0;i<=n;i++)	g[i]=inv[i];
	for(m=n;(1<<len)<=n+n;++len);
	n=1<<len,Inv=fsp(n,P-2),reverse(f,f+n);
	for(int i=1;i<n;i++)	r[i]=(r[i>>1]>>1)|((i&1)<<(len-1));
	NTT(f,0),NTT(g,0);
	for(int i=0;i<n;i++)	f[i]=f[i]*g[i]%P;
	NTT(f,1),reverse(f,f+n);
	for(int i=1;i<=m;i++)	printf("%lld\n",(m*C(m,i)-f[i]*inv[i]%P+P)%P);
	return 0;
}
