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=(1<<20)+5;
int lim=1,M=924844033,a[N],r[N],fac[N],inv[N],b[N],tot,h[N],sz[N],g=5,gi=554906420,n,i,x,y,l;
struct node{
	int to,ne;
}e[N];
inline int mul(int x,int y){return 1ll*x*y%M;}
inline int inc(int x,int y){x+=y;if(x>=M)x-=M;return x;}
inline int dec(int x,int y){x-=y;if(x<0)x+=M;return x;}
void add(int x,int y){e[++tot]={y,h[x]},h[x]=tot;}
int pw(int x,int y){
	int z=1;
	for (;y;y>>=1,x=1ll*x*x%M)
		if (y&1) z=1ll*z*x%M;
	return z;
}
int Inv(int x){return pw(x,M-2);}
void dfs(int u,int fa){
	sz[u]=1;
	for (int i=h[u],v;i;i=e[i].ne)
		if ((v=e[i].to)!=fa) dfs(v,u),sz[u]+=sz[v],a[sz[v]]++;
	a[n-sz[u]]++;
}
void ntt(int *a,int op){
	for (int i=0;i<lim;i++)
		if (i<r[i]) swap(a[i],a[r[i]]);
	for (int mid=1,pp=M-1>>1;mid<lim;mid<<=1,pp>>=1)
		for (int R=mid<<1,j=0,wn=pw(op==1?g:gi,pp);j<lim;j+=R)
			for (int k=0,w=1;k<mid;k++,w=mul(w,wn)){
				int x=a[j|k],y=mul(w,a[j|k|mid]);
				a[j|k]=inc(x,y);
				a[j|k|mid]=dec(x,y);
			}
	if (op==-1){
		int inv=Inv(lim);
		for (int i=0;i<lim;i++) a[i]=mul(a[i],inv);
	}
}
int main(){
	scanf("%d",&n);
	for (i=1;i<n;i++) scanf("%d%d",&x,&y),add(x,y),add(y,x);
	dfs(1,0);
	fac[0]=1,a[0]=0;
	for (i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%M,a[i]=1ll*a[i]*fac[i]%M;
	b[0]=inv[n]=Inv(fac[n]);
	for (i=n-1;~i;i--) inv[i]=1ll*inv[i+1]*(i+1)%M,b[n-i]=inv[i];
	for (;lim<=n+n;lim<<=1,l++);
	for (i=0;i<lim;i++) r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
	ntt(a,1),ntt(b,1);
	for (i=0;i<lim;i++) a[i]=1ll*a[i]*b[i]%M;
	ntt(a,-1);
	for (i=1;i<=n;i++) printf("%lld\n",(1ll*fac[n]*inv[i]%M*inv[n-i]%M*n%M-1ll*inv[i]*a[n+i]%M+M)%M);
}