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<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<vector>
#include<iostream>
#include<set>
#include<map>
#include<functional>
#include<queue>
#include<string>

using namespace std;

typedef long long LL;
const int MX=524299,md=924844033;
int g;
int w[2][MX],rev[MX];
int n,N;
vector<int>G[MX];

inline int pw(int x,int y){
	int ans=1;
	while(y){
		if(y&1)ans=(LL)ans*x%md;
		x=(LL)x*x%md,y>>=1;
	}
	return ans;
}
inline int inv(int x){return pw(x,md-2);}
bool ck(int g){
	for(int i=2;i*i<=md-1;i++)if((md-1)%i==0&&(pw(g,i)==1||pw(g,(md-1)/i)==1))return false;
	return true;
}
void calg(){
	g=2;while(!ck(g))++g;
}
void ini(){
	w[0][0]=1;
	int G=pw(g,(md-1)/N);
	for(int i=1;i<N;i++)w[0][i]=(LL)w[0][i-1]*G%md;
	for(int i=0;i<N;i++)w[1][i]=w[0][(N-i)%N];
	for(int i=1;i<N;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(N>>1));
}
void FFT(int a[],int typ){
	for(int i=0;i<N;i++)if(i<rev[i])swap(a[i],a[rev[i]]);
	int x,y;
	for(int i=1;i<N;i<<=1)
		for(int j=0,d=N/(i<<1);j<N;j+=(i<<1))
			for(int k=j,l=0;k<j+i;k++,l+=d)
				x=(LL)w[typ][l]*a[k+i]%md,y=a[k],a[k]=(y+x)%md,a[k+i]=(md+y-x)%md;
	if(typ){
		int iv=inv(N);
		for(int i=0;i<N;i++)a[i]=(LL)a[i]*iv%md;
	}
}

int c[MX];

int dfs(int k,int f){
	int sz=1;
	for(vector<int>::iterator it=G[k].begin();it!=G[k].end();it++)if(*it!=f)sz+=dfs(*it,k);
	if(k!=1)++c[sz],++c[n-sz];
	return sz;
}
int main(){
	calg();
	scanf("%d",&n);
	for(N=1;N<n;N<<=1);N<<=1;
	ini();
	for(int i=1;i<n;i++){
		static int u,v;
		scanf("%d%d",&u,&v);
		G[u].push_back(v),G[v].push_back(u);
	}
	dfs(1,0);
	static int fac[MX],facinv[MX],inv[MX];
	inv[1]=1;
	for(int i=2;i<=n;i++)inv[i]=(LL)inv[md%i]*(md-md/i)%md;
	fac[0]=facinv[0]=1;
	for(int i=1;i<=n;i++)fac[i]=(LL)fac[i-1]*i%md;
	for(int i=1;i<=n;i++)facinv[i]=(LL)facinv[i-1]*inv[i]%md;
	
	static int a[MX],b[MX];
	for(int i=0;i<n;i++)a[i]=facinv[i];
	for(int i=0;i<n;i++)b[i]=(LL)c[n-i]*fac[n-i]%md;
	FFT(a,0),FFT(b,0);
	for(int i=0;i<N;i++)a[i]=(LL)a[i]*b[i]%md;
	FFT(a,1);
	for(int i=1;i<=n;i++)printf("%lld\n",((LL)n*fac[n]%md*facinv[i]%md*facinv[n-i]%md-(LL)facinv[i]*a[n-i]%md+md)%md); 
	return 0;
}