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"

using namespace std;

const int maxm=1e6+10;
const long long Mod=924844033;
const long long G=5;

int head[maxm],tot,N;
int Father[maxm],Size[maxm],Limit=1,l;
long long Inv1[maxm],Inv2[maxm],Fac[maxm],r[maxm];
long long Cnt[maxm],A[maxm<<2],B[maxm<<2];

struct edge{
	int To;
	int Next;
	# define To(x) Edge[x].To
	# define Next(x) Edge[x].Next
}Edge[maxm<<1];

inline void Add_Edge(int From,int To){Edge[++tot]=(edge){To,head[From]},head[From]=tot;return;}
inline void Insert_Edge(int x,int y){Add_Edge(x,y),Add_Edge(y,x);return;}

void Dfs(int now){
	Size[now]=1;
	for(int i=head[now];i;i=Next(i)){
		if(To(i)==Father[now]) continue;
		Father[To(i)]=now;
		Dfs(To(i));
		Size[now]+=Size[To(i)];
		Cnt[Size[To(i)]]--;
	}
	Cnt[N-Size[now]]--;
	return;
}

inline long long power(long long x,long long y){
	long long Ans=1;
	while(y){
		if(y&1) Ans=Ans*x%Mod;
		x=x*x%Mod;
		y>>=1;
	}
	return Ans;
}

void NTT(long long *A,int n,int type){
	register int i,j,k,mid,R;
	register long long Wn,W,x,y;
	for(i=0;i<n;i++) if(i<r[i]) swap(A[i],A[r[i]]);
	for(mid=1;mid<n;mid<<=1){
		Wn=power(G,(Mod-1)/(mid<<1));
		if(type==-1) Wn=power(Wn,Mod-2);
		for(R=mid<<1,j=0;j<n;j+=R){
			W=1;
			for(k=0;k<mid;k++,W=W*Wn%Mod){
				x=A[j+k],y=W*A[j+mid+k]%Mod;
				A[j+k]=(x+y)%Mod;
				A[j+k+mid]=(x-y+Mod)%Mod;
			}
		}
	}
	if(type==-1){
		int Inv=power(n,Mod-2);
		for(i=0;i<=n;i++) A[i]=A[i]*Inv%Mod;
	}
	return;
}

inline void Mul(long long *A,long long *B,int N){
	NTT(A,N,1);NTT(B,N,1);
	for(register int i=0;i<=N;i++) A[i]=A[i]*B[i]%Mod;
	NTT(A,N,-1);
	return;
}

int main(){
	register int i;
	register long long Now;
	int x,y;
	scanf("%d",&N);
	Inv1[1]=Inv2[1]=Fac[1]=Inv1[0]=Inv2[0]=Fac[0]=1;
	for(i=2;i<=N;i++){
		scanf("%d%d",&x,&y);
		Insert_Edge(x,y);
		Fac[i]=Fac[i-1]*i%Mod;
		Inv1[i]=(Mod-Mod/i)*Inv1[Mod%i]%Mod;
		Inv2[i]=Inv1[i]*Inv2[i-1]%Mod;
	}
	Dfs(1);
	for(i=1;i<=N;i++){
        Cnt[i]=(Mod+Cnt[i])%Mod;
        A[i]=Cnt[i]*Fac[i]%Mod;
        B[i]=Inv2[N-i];
	}
	A[N]=Fac[N]*N%Mod;
	while(Limit<=(N<<1)) Limit<<=1,l++;
	for(i=0;i<Limit;i++) r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
	Mul(A,B,Limit);
	for(i=1;i<=N;i++){
		Now=A[N+i]*Inv2[i]%Mod;
		Now=(Now+Mod)%Mod;
		printf("%lld\n",A[N+i]*Inv2[i]%Mod);
	}
	return 0;
}