Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N=8e5+5,P=924844033;
int n,hd[N],V[N],nx[N],t,p,A[N],G[2]={5,554906420},B[N],sz[N],r[N],jc[N],ny[N];
int X(int x){return x>=P?x-P:x;}
int K(int x,int y){
	int z=1;
	for (;y;y>>=1,x=1ll*x*x%P)
		if (y&1) z=1ll*z*x%P;
	return z;
}
void add(int u,int v){
	nx[++t]=hd[u];V[hd[u]=t]=v;
}
void dfs(int u,int fr){
	sz[u]=1;
	for (int i=hd[u];i;i=nx[i])
		if (V[i]!=fr){
			dfs(V[i],u);
			sz[u]+=sz[V[i]];
			A[sz[V[i]]]--;
		}
	if (u!=1) A[n-sz[u]]--;
}
void Ntt(int *a,int o){
	for (int i=0;i<t;i++)
		if (i<r[i]) swap(a[i],a[r[i]]);
	for (int wn,i=1;i<t;i<<=1){
		wn=K(G[o],(P-1)/(i<<1));
		for (int x,y,j=0;j<t;j+=(i<<1))
			for (int k=0,w=1;k<i;k++,w=1ll*w*wn%P)
				x=a[j+k],y=1ll*a[i+j+k]*w%P,
				a[j+k]=X(x+y),a[i+j+k]=X(x-y+P);
	}
	if (o)
		for (int i=0,v=K(t,P-2);i<t;i++)
			a[i]=1ll*a[i]*v%P;
}
int main(){
	cin>>n;jc[0]=1;
	for (int i=1,x,y;i<n;i++)
		scanf("%d%d",&x,&y),
		add(x,y),add(y,x);
	A[n]=n;dfs(1,0);
	for (int i=1;i<=n;i++)
		jc[i]=1ll*i*jc[i-1]%P;
	ny[n]=K(jc[n],P-2);
	for (int i=n;i;i--)
		ny[i-1]=1ll*i*ny[i]%P;
	for (int i=0;i<=n;i++)
		B[i]=ny[i],A[i]=1ll*X(A[i]+P)*jc[i]%P;
	reverse(B,B+n+1);
	for (t=1;t<n+n+2;t<<=1,p++);
	for (int i=0;i<t;i++)
		r[i]=(r[i>>1]>>1)|((i&1)<<(p-1));
	Ntt(A,0);Ntt(B,0);
	for (int i=0;i<t;i++) A[i]=1ll*A[i]*B[i]%P;
	Ntt(A,1);
	for (int i=1;i<=n;i++) printf("%lld\n",1ll*ny[i]*A[n+i]%P);
	return 0;
}