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<algorithm>
using namespace std;
inline int read()
{
	int x;char c;
	while((c=getchar())<'0'||c>'9');
	for(x=c-'0';(c=getchar())>='0'&&c<='9';)x=x*10+c-'0';
	return x;
}
#define N 524288
#define MOD 924844033
struct edge{int nx,t;}e[N+5];
int h[N+5],en,s[N+5],w[2][N+5],R[N+5],f[N+5],rf[N+5],a[N+5],b[N+5];
inline void ins(int x,int y)
{
	e[++en]=(edge){h[x],y};h[x]=en;
	e[++en]=(edge){h[y],x};h[y]=en;
}
inline int pw(int x,int y)
{
	int r=1;
	for(;y;y>>=1,x=1LL*x*x%MOD)if(y&1)r=1LL*r*x%MOD;
	return r;
}
void dfs(int x,int fa)
{
	s[x]=1;
	for(int i=h[x];i;i=e[i].nx)if(e[i].t!=fa)
		dfs(e[i].t,x),s[x]+=s[e[i].t];
}
void init()
{
	int g=pw(5,(MOD-1)/N),i,j,k;
	for(w[0][0]=i=1;i<N;++i)w[0][i]=1LL*w[0][i-1]*g%MOD;
	for(w[1][0]=i=1;i<N;++i)w[1][i]=w[0][N-i];
	for(i=j=0;i<N;R[++i]=j)for(k=N>>1;(j^=k)<k;k>>=1);
}
inline int mod(int x){return x<MOD?x:x-MOD;}
void fft(int*x,int v)
{
	int i,j,k,p;
	for(i=0;i<N;++i)if(i<R[i])swap(x[i],x[R[i]]);
	for(i=1;i<N;i<<=1)for(j=0;j<N;j+=i<<1)for(k=0;k<i;++k)
	{
		p=1LL*x[i+j+k]*w[v][N/(i<<1)*k]%MOD;
		x[i+j+k]=mod(x[j+k]+MOD-p);x[j+k]=mod(x[j+k]+p);
	}
	if(v)for(i=0,p=pw(N,MOD-2);i<N;++i)x[i]=1LL*x[i]*p%MOD;
}
int main()
{
	int n=read(),i; 
	for(i=1;i<n;++i)ins(read(),read());
	dfs(1,0);
	for(i=2;i<=n;++i)++a[s[i]],++a[n-s[i]];
	for(i=f[0]=1;i<=n;++i)f[i]=1LL*f[i-1]*i%MOD;
	for(rf[i=n]=pw(f[n],MOD-2);i--;)rf[i]=1LL*rf[i+1]*(i+1)%MOD;
	for(i=1;i<=n;++i)a[i]=1LL*a[i]*f[i]%MOD,b[i]=rf[n-i];
	init();fft(a,0);fft(b,0);
	for(i=0;i<N;++i)a[i]=1LL*a[i]*b[i]%MOD;
	fft(a,1);
	for(i=1;i<=n;++i)printf("%d\n",(1LL*n*f[n]%MOD*rf[i]%MOD*rf[n-i]+MOD-1LL*a[i+n]*rf[i]%MOD)%MOD);
}