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<bits/stdc++.h>
typedef long long i64;
const int P=924844033,M=2e5+7;
int pw(int a,int n){
	if(n<0)n+=P-1;
	int v=1;
	for(;n;n>>=1,a=i64(a)*a%P)if(n&1)v=i64(v)*a%P;
	return v;
}
int A[533333],B[533333],r[533333],E[533333];
int n,ep=2,fac[M],fiv[M],e0[M],sz[M],t[M],N,K;
struct edge{
	int to,nx;
}e[M*2];
void f1(int w,int pa){
	sz[w]=1;
	for(int i=e0[w];i;i=e[i].nx){
		int u=e[i].to;
		if(u!=pa)f1(u,w),sz[w]+=sz[u];
	}
	++t[sz[w]];
	++t[n-sz[w]];
}
int fix(int x){return x+(x>>31&P);}
void ntt(int*a,int t){
	for(int i=1;i<N;++i)if(i<r[i])std::swap(a[i],a[r[i]]);
	for(int i=1;i<N;i<<=1){
		for(int j=0;j<N;j+=i<<1){
			i64 z=pw(5,(P-1)/(i<<1)*t);
			E[0]=1;
			for(int k=1;k<i;++k)E[k]=E[k-1]*z%P;
			int*b=a+j,*c=b+i;
			for(int k=0;k<i;++k){
				int x=b[k],y=c[k]*i64(E[k])%P;
				b[k]=fix(x+y-P);
				c[k]=fix(x-y);
			}
		}
	}
	if(t==-1){
		i64 I=pw(N,-1);
		for(int i=0;i<N;++i)a[i]=a[i]*I%P;
	}
}
int main(){
	scanf("%d",&n);
	for(int i=fac[0]=1;i<=n;++i)fac[i]=i64(i)*fac[i-1]%P;
	fiv[n]=pw(fac[n],-1);
	for(int i=n;i;--i)fiv[i-1]=i64(i)*fiv[i]%P;
	for(int i=1,a,b;i<n;++i){
		scanf("%d%d",&a,&b);
		e[ep]=(edge){b,e0[a]};e0[a]=ep++;
		e[ep]=(edge){a,e0[b]};e0[b]=ep++;
	}
	f1(1,0);
	for(N=2,K=0;N<n*2+5;N<<=1,++K);
	for(int i=1;i<N;++i)r[i]=r[i>>1]>>1|(i&1)<<K;
	for(int i=1;i<n;++i)A[n-i]=i64(t[i])*fac[i]%P;
	for(int i=0;i<=n;++i)B[i]=fiv[i];
	ntt(A,1),ntt(B,1);
	for(int i=0;i<N;++i)A[i]=i64(A[i])*B[i]%P;
	ntt(A,-1);
	for(int i=1;i<=n;++i){
		int v=(i64(n)*fac[n]%P*fiv[n-i]%P-A[n-i])*fiv[i]%P;
		printf("%d\n",(v+P)%P);
	}
	return 0;
}