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>
#define maxn 530100
#define mod 924844033
using namespace std;
struct edge{int r,nxt;}e[maxn<<1];
int n,esz,head[maxn],siz[maxn],h[maxn],fac[maxn],inv[maxn];
int anses[maxn],A[maxn],B[maxn],bh[maxn],C[maxn],D[maxn];
int qpow(int a,int b){
	int ans=1,tmp=a;
	for(;b;b>>=1,tmp=1ll*tmp*tmp%mod)
		if(b&1)ans=1ll*ans*tmp%mod;
	return ans;
}
void addedge(int u,int v){
	e[++esz].r=v;e[esz].nxt=head[u];head[u]=esz;
	e[++esz].r=u;e[esz].nxt=head[v];head[v]=esz;
}
void dfs(int u,int f){
	siz[u]=1;
	for(int t=head[u];t;t=e[t].nxt)if(e[t].r!=f)
		dfs(e[t].r,u),siz[u]+=siz[e[t].r];
}
int getC(int x,int y){
	if(x>y||x<0)return 0;
	return 1ll*fac[y]*inv[x]%mod*inv[y-x]%mod;
}
void fft(int h[],int len,int flag){
	for(int i=0;i<len;++i)if(bh[i]>i)swap(h[i],h[bh[i]]);
	for(int i=1;i<len;i<<=1){
		int wn=qpow(5,(flag==-1?(mod-1)-(mod-1)/(i<<1):(mod-1)/(i<<1)));
		for(int j=0;j<len;j+=(i<<1)){
			int w=1;
			for(int k=0;k<i;++k){
				int x=h[j+k],y=1ll*h[j+k+i]*w%mod;
				h[j+k]=(x+y)%mod;
				h[j+k+i]=(x-y+mod)%mod;
				w=1ll*w*wn%mod;
			}
		}
	}
	if(flag==-1){
		int inv=qpow(len,mod-2);
		for(int i=0;i<len;++i)
			h[i]=1ll*h[i]*inv%mod;
	}
}
int main(){
	scanf("%d",&n),fac[0]=inv[0]=1;
	for(int i=1;i<=n;++i)
		fac[i]=1ll*fac[i-1]*i%mod;
	inv[n]=qpow(fac[n],mod-2);
//	if(1ll*inv[n]*fac[n]%mod!=1)exit(-1);
	for(int i=n-1;i>=1;--i)
		inv[i]=1ll*inv[i+1]*(i+1)%mod;
	for(int i=2,u,v;i<=n;++i)
		scanf("%d%d",&u,&v),addedge(u,v);
	dfs(1,0);
	for(int i=1;i<=n;++i)anses[i]=1ll*(n+1)*getC(i,n)%mod;
	for(int i=1;i<=n;++i)A[siz[i]]++;
	int len=1,k=0;
	while(len<=n*2)len<<=1,k++;
	for(int i=0;i<len;i++)bh[i]=(bh[i>>1]>>1)|((i&1)<<(k-1));
	for(int i=0;i<=n;++i)B[i]=1ll*A[i]*fac[i]%mod;
	for(int i=0;i<=n;++i)C[i]=inv[n-i];
	fft(B,len,1),fft(C,len,1);
	for(int i=0;i<len;++i)B[i]=1ll*B[i]*C[i]%mod;
	fft(B,len,-1);
	for(int i=1;i<=n;++i)anses[i]=(anses[i]-1ll*B[i+n]*inv[i])%mod;
	memset(B,0,sizeof(B)),memset(C,0,sizeof(C));
	for(int i=0;i<=n;++i)B[i]=1ll*A[i]*fac[n-i]%mod;
	for(int i=0;i<=n;++i)C[i]=inv[i];
	fft(B,len,1),fft(C,len,1);
	for(int i=0;i<len;++i)B[i]=1ll*B[i]*C[i]%mod;
	fft(B,len,-1);
	for(int i=1;i<=n;++i)anses[i]=(anses[i]-1ll*B[n-i]*inv[i])%mod;
	for(int i=1;i<=n;++i)printf("%d\n",(anses[i]%mod+mod)%mod);
}