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
/*ID:PROG:LANG:*/
#include<bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f
#define MOD 924844033
#define MAXN 200005
using namespace std;
struct Edge{
	int to,nex;Edge(int _=0,int __=0){to=_,nex=__;}
}e[MAXN<<1];
const LL g=5,gi=554906420;
LL n,cnt,mx=1,len,hd[MAXN],jc[MAXN],njc[MAXN],sz[MAXN],num[MAXN],f[MAXN],r[MAXN<<2],a[MAXN<<2],b[MAXN<<2];
LL rd(){
	LL x=0,tp=1;char c=getchar();
	while(c<'0' || c>'9'){if(c=='-')tp=-1;c=getchar();}
	while(c>='0' && c<='9'){x=(x*10+c-'0');c=getchar();}
	return x*tp;
}
void adde(int x,int y){
	e[++cnt]=Edge(y,hd[x]);
	hd[x]=cnt;
}
LL quick_mod(LL a,LL x){LL res=1;while(x){if(x&1)(res*=a)%=MOD;(a*=a)%=MOD,x>>=1;}return res;}
LL C(int n,int m){if(n<m)return 0;return jc[n]*njc[m]%MOD*njc[n-m]%MOD;}
void ntt(LL *a,LL n,LL tp){
	for(int i=0;i<mx;i++)if(i<r[i])swap(a[i],a[r[i]]);
	for(int mid=1;mid<n;mid<<=1){
		LL Wn=quick_mod((tp==1)?g:gi,(MOD-1)/(mid<<1));
		for(int i=0;i<n;i+=mid<<1){
			LL w=1;
			for(int j=0;j<mid;j++){
				LL x=a[i+j],y=w*a[i+j+mid]%MOD;
				a[i+j]=(x+y)%MOD,a[i+j+mid]=(x-y+MOD)%MOD;
				(w*=Wn)%=MOD;
			}
		}
	}
	LL inv=quick_mod(n,MOD-2);
	if(tp==-1)for(int i=0;i<mx;i++)(a[i]*=inv)%=MOD;
}
void dfs(int now,int fa){
	sz[now]=1;
	for(int i=hd[now];i;i=e[i].nex)
		if(e[i].to!=fa)
			dfs(e[i].to,now),sz[now]+=sz[e[i].to];
	if(fa)num[sz[now]]++,num[n-sz[now]]++;
}
int main(){
	// freopen("testdata.in","r",stdin);
	// freopen("1.out","w",stdout);
	int x,y;jc[0]=njc[0]=1;
	n=rd();
	for(int i=1;i<=n;i++)jc[i]=jc[i-1]*i%MOD;
	njc[n]=quick_mod(jc[n],MOD-2);
	for(int i=n-1;i>=1;i--)njc[i]=njc[i+1]*(i+1)%MOD;
	for(int i=1;i<n;i++){
		x=rd(),y=rd();
		adde(x,y),adde(y,x);
	}
	dfs(1,0);
	while(mx<=(n<<1))mx<<=1,len++;
	for(int i=1;i<mx;i++)r[i]=(r[i>>1]>>1)|((i&1)<<len-1);
	for(int i=1;i<=n;i++)a[i]=num[i]*jc[i]%MOD;
	for(int i=0;i<=n;i++)b[n-i]=njc[i];
	ntt(a,mx,1),ntt(b,mx,1);
	for(int i=0;i<mx;i++)(a[i]*=b[i])%=MOD;
	ntt(a,mx,-1);
	for(int i=1;i<=n;i++)f[i]=a[n+i]*njc[i]%MOD;
	for(int i=1;i<=n;i++)printf("%lld\n",(n*C(n,i)%MOD-f[i]+MOD)%MOD);
	// printf("%.2lf\n",(double)clock()/CLOCKS_PER_SEC);
	return 0;
}