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:
// ===================================
//   author: M_sea
//   website: http://m-sea-blog.com/
// ===================================
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#define register
using namespace std;

inline int read()
{
	int X=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c>='0'&&c<='9') X=X*10+c-'0',c=getchar();
	return X*w;
}

const int MAXN=600000+10;
const int mod=924844033;
inline int qpow(int a,int b)
{
	int c=1;
	for(; b; b>>=1,a=1ll*a*a%mod) if(b&1) c=1ll*c*a%mod;
	return c;
}
int fac[MAXN],ifac[MAXN];
int cnt[MAXN],sz[MAXN];
int rev[MAXN],f[MAXN],g[MAXN];
int n;

struct edge {int v,nxt;} e[MAXN<<1];
int head[MAXN];
inline void addEdge(int u,int v)
{
	static int c=0;
	e[++c]=(edge) {v,head[u]},head[u]=c;
}


inline void init(int n)
{
	fac[0]=1;
	for(int i=1; i<=n; ++i) fac[i]=1ll*fac[i-1]*i%mod;
	ifac[n]=qpow(fac[n],mod-2);
	for(int i=n; i; --i) ifac[i-1]=1ll*ifac[i]*i%mod;
}

inline int C(int n,int m)
{return 1ll*fac[n]*ifac[m]%mod*ifac[n-m]%mod;}


inline void dfs(int u,int fa)
{
	sz[u]=1;
	for(int i=head[u]; i; i=e[i].nxt)
	{
		int v=e[i].v; if(v==fa) continue;
		dfs(v,u),sz[u]+=sz[v],++cnt[sz[v]];
	}
	++cnt[n-sz[u]];
}

inline void ntt(int *a,int n,int op)
{
	for(int i=0; i<n; ++i) if(i<rev[i]) swap(a[i],a[rev[i]]);
	for(int i=1; i<n; i<<=1)
	{
		int rot=qpow(op==1?5:554906420,(mod-1)/(i<<1));
		for(int j=0; j<n; j+=i<<1)
			for(int k=0,w=1; k<i; ++k,w=1ll*w*rot%mod)
			{
				int x=a[j+k],y=1ll*w*a[j+k+i]%mod;
				a[j+k]=(x+y)%mod,a[j+k+i]=(x-y+mod)%mod;
			}
	}
	if(op==-1)
	{
		int inv=qpow(n,mod-2);
		for(int i=0; i<n; ++i) a[i]=1ll*a[i]*inv%mod;
	}
}

int main()
{
	n=read();
	for(int i=1; i<n; ++i)
	{
		int u=read(),v=read();
		addEdge(u,v),addEdge(v,u);
	}
	dfs(1,0); init(n);
	for(int i=1; i<=n; ++i) f[i]=1ll*cnt[i]*fac[i]%mod;
	reverse(f,f+n+1);
	for(int i=0; i<=n; ++i) g[i]=ifac[i];
	int lim=1,l=0;
	for(; lim<=n<<1; lim<<=1,++l);
	for(int i=0; i<lim; ++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
	ntt(f,lim,1),ntt(g,lim,1);
	for(int i=0; i<lim; ++i) f[i]=1ll*f[i]*g[i]%mod;
	ntt(f,lim,-1);
	reverse(f,f+n+1);
	for(int i=1; i<=n; ++i)
		printf("%lld\n",(1ll*n*C(n,i)%mod
		                 -1ll*ifac[i]*f[i]%mod+mod)%mod);
	return 0;
}