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 <iostream>
#include <cstdio>
#include <cstring>
#define N 200005
using namespace std;
typedef long long ll;
const int mod=924844033;
struct edge{
	int k,next;
}e[N<<1];
int n,home[N],cnt=-1;
ll f[N],inv[N],num[N];
ll quick_pow(ll a,ll b){
	ll ans=1;
	for(;b;b>>=1,a=a*a%mod)
		if(b&1) ans=ans*a%mod;
	return ans;
}
void add(int x,int y){
	cnt++;
	e[cnt].k=y;
	e[cnt].next=home[x];
	home[x]=cnt;
}
int siz[N],m;
void dft(ll *a,int sig){
	for(int i=0;i<m;i++){
		int p=0,g=m>>1,j=i;
		while(g) p=p<<1|(j&1),g>>=1,j>>=1;
		if(p<i) swap(a[p],a[i]);
	}
	ll w,wn;
	for(int b=2;b<=m;b<<=1){
		wn=quick_pow(5,sig==1?(mod-1)/b:mod-1-(mod-1)/b);
		int half=b>>1;
		for(int i=0;i<m;i+=b){
			w=1;
			for(int j=0;j<half;j++){
				ll u=a[i+j],v=a[i+j+half]*w%mod;
				a[i+j]=(u+v)%mod;
				a[i+j+half]=(u-v)%mod;
				w=w*wn%mod;
			}
		}
	}
	ll inv=quick_pow(m,mod-2);
	if(sig==-1) for(int i=0;i<m;i++) a[i]=a[i]*inv%mod;
}
void dfs(int k,int F){
	siz[k]=1;
	for(int i=home[k];~i;i=e[i].next){
		if(e[i].k!=F){
			dfs(e[i].k,k);
			siz[k]+=siz[e[i].k];
		}
	}
	num[siz[k]]--;
	num[n-siz[k]]--;
}
ll a[N<<2],b[N<<2];
int main(){
	memset(home,-1,sizeof(home));
	scanf("%d",&n);
	for(int i=1,x,y;i<n;i++){
		scanf("%d%d",&x,&y);
		add(x,y);add(y,x);
	}
	f[0]=inv[0]=1;
	for(int i=1;i<=n;i++) f[i]=f[i-1]*i%mod,inv[i]=quick_pow(f[i],mod-2);
	dfs(1,0);
	m=1;
	while(m<=n+n) m<<=1;
	num[n]=n;
	b[0]=1;
	for(int i=1;i<=n;i++) a[n-i]=num[i]*f[i]%mod,b[i]=inv[i];
	dft(a,1);
	dft(b,1);
	for(int i=0;i<m;i++) a[i]=a[i]*b[i]%mod;
	dft(a,-1);
	for(int i=1;i<=n;i++) printf("%lld\n",(a[n-i]+mod)*inv[i]%mod);
	return 0;
}