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 <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 800010
#define LL long long
#define P 924844033
using namespace std;

LL getPow(LL x,LL y){
	LL res=1;
	while(y){
		if(y&1) res=res*x%P;
		x=x*x%P;
		y>>=1;
	}
	return res;
}

struct edge{
	int to,next;
	edge(int _to=0,int _next=0):to(_to),next(_next){}
}e[MAXN<<1];

int n,sizew;
int g[MAXN],nume;
int size[MAXN];
LL fac[MAXN],invfac[MAXN];
LL c[MAXN],d[MAXN],f[MAXN];

void addEdge(int u,int v){
	e[nume]=edge(v,g[u]);
	g[u]=nume++;
}

void dfs(int x,int p){
	size[x]=1;
	for(int i=g[x];~i;i=e[i].next)
		if(e[i].to^p){
			dfs(e[i].to,x);
			size[x]+=size[e[i].to];
		}
}

void init(){
	fac[0]=1;
	for(int i=1;i<MAXN;i++) fac[i]=fac[i-1]*i%P;
	invfac[MAXN-1]=getPow(fac[MAXN-1],P-2);
	for(int i=MAXN-2;i>=0;i--) invfac[i]=invfac[i+1]*(i+1)%P;
}

LL getC(int x,int y){
	if(x<y) return 0;
	return fac[x]*invfac[y]%P*invfac[x-y]%P;
}

void FFT(LL *a,int len,int flag){
	static int rev[MAXN];
	rev[0]=0;
	for(int i=1;i<len;i++) rev[i]=rev[i>>1]>>1|((i&1)?(len>>1):0);
	for(int i=0;i<len;i++)
		if(i<rev[i]) 
			swap(a[i],a[rev[i]]);
	for(int l=2;l<=len;l<<=1){
		LL w=getPow(5,(P-1)/l);
		int l2=l/2;
		for(int i=0;i<len;i+=l){
			LL temp=1;
			for(int j=0;j<l2;j++){
				LL t1=a[i+j],t2=a[i+j+l2]*temp%P;
				a[i+j]=(t1+t2)%P;
				a[i+j+l2]=(t1-t2)%P;
				temp=temp*w%P;
			}
		}
	}
	if(flag==-1){
		LL invn=getPow(len,P-2);
		for(int i=0;i<len;i++)
			a[i]=a[i]*invn%P;
		for(int i=1;i<len;i++)
			if(i<len-i)
				swap(a[i],a[len-i]);
	}
}

int main(){
#ifdef DEBUG
	freopen("F.in","r",stdin);
#endif
	init();
	memset(g,-1,sizeof g);
	scanf("%d",&n);
	for(sizew=1;sizew<=n;sizew<<=1);
	sizew<<=1;
	for(int i=1;i<n;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		addEdge(u,v);
		addEdge(v,u);
	}
	dfs(1,0);
	for(int i=2;i<=n;i++){
		c[size[i]]++;
		c[n-size[i]]++;
	}
	for(int i=1;i<=n;i++) c[i]=c[i]*fac[i]%P;
	for(int i=0;i<=n;i++) d[i]=invfac[n-i];
	FFT(c,sizew,1); FFT(d,sizew,1);
	for(int i=0;i<sizew;i++) f[i]=c[i]*d[i]%P;
	FFT(f,sizew,-1);
	for(int i=1;i<=n;i++){
		f[i]=f[i+n]*invfac[i]%P;
		f[i]=((getC(n,i)*n-f[i])%P+P)%P;
		printf("%lld\n",f[i]);
	}
	return 0;
}
