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 <bits/stdc++.h>
using namespace std;

inline int read()
{
	int x=0,f=1,c=getchar();
	while(c<48) c=='-'&&(f=-1),c=getchar();
	while(c>47) x=x*10+c-'0',c=getchar();
	return x*f;
}

const int MAXN = 600005;
const int mod = 924844033;
std::vector<int> G[MAXN];
int fac[MAXN],ifac[MAXN],cnt[MAXN],sz[MAXN];
int f[MAXN],g[MAXN],rev[MAXN];
int n;

inline void addedge(int u,int v)
{
	G[u].push_back(v);
	G[v].push_back(u);
}

inline int qpow(int x,int k)
{
	int res=1;
	for(int i=k; i; i>>=1,x=1ll*x*x%mod)
		if(i&1) res=1ll*res*x%mod;
	return res;
}

inline void init(int n)
{
	for(int i=fac[0]=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 cat(int n,int k)
{return 1ll*fac[n]*ifac[k]%mod*ifac[n-k]%mod;}

inline void dfs(int x,int fa)
{
	sz[x]=1;
	for(int y : G[x])
		if(y^fa) dfs(y,x),sz[x]+=sz[y],++cnt[sz[y]];
	++cnt[n-sz[x]];
}

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(int argc, char const *argv[])
{
	n=read();
	for(int i=1; i<n; ++i) addedge(read(),read());
	dfs(1,0); init(n);
	for(int i=1; i<=n; ++i) f[i]=1ll*cnt[i]*fac[i]%mod;
	for(int i=(reverse(f,f+n+1),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*cat(n,i)%mod
		                 -1ll*ifac[i]*f[i]%mod+mod)%mod);
	return 0;
}