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;

typedef long long ll;
const int maxn=1<<18,mod=924844033;
int n,size[maxn];
ll fac[maxn],finv[maxn],a[maxn<<1],b[maxn<<1],c[maxn<<1];
vector<int> g[maxn];

inline ll qpow(ll a,ll n){
	ll res=1;
	while(n){
		if(n&1)
			res=res*a%mod;
		a=a*a%mod;
		n>>=1;
	}
	return res;
}
void dfs(int pos,int fa){
	size[pos]=1;
	for(int i=0,v;i<g[pos].size();++i)
		if((v=g[pos][i])!=fa){
			dfs(v,pos);
			size[pos]+=size[v];
			++a[size[v]];
			++a[n-size[v]];
		}
}
inline int rev(int id,int len){
	int res=0;
	for(int i=1;i<len;i<<=1){
		res<<=1;
		if(id&i)
			res|=1;
	}
	return res;
}
inline void fft(ll*a,int len,int dft){
	for(int i=0;i<len;++i)
		if(i<rev(i,len))
			swap(a[i],a[rev(i,len)]);
	for(int i=2;i<=len;i<<=1){
		ll w=qpow(5,(mod-1)/i);
		if(dft==-1)
			w=qpow(w,mod-2);
		for(int j=0;j<len;j+=i){
			ll ww=1;
			for(int k=0;k<(i>>1);++k,ww=ww*w%mod){
				ll x=a[j+k],y=a[j+k+(i>>1)]*ww%mod;
				a[j+k]=(x+y)%mod;
				a[j+k+(i>>1)]=(x-y+mod)%mod;
			}
		}
	}
	if(dft==-1){
		int inv=qpow(len,mod-2);
		for(int i=0;i<len;++i)
			a[i]=a[i]*inv%mod;
	}
}

int main(){
	fac[0]=1;
	for(int i=1;i<maxn;++i)
		fac[i]=fac[i-1]*i%mod;
	finv[maxn-1]=qpow(fac[maxn-1],mod-2);
	for(int i=maxn-1;i;--i)
		finv[i-1]=finv[i]*i%mod;
	scanf("%d",&n);
	for(int i=1,u,v;i<n;++i){
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(1,0);
	for(int i=1;i<n;++i)
		a[i]=a[i]*fac[i]%mod;
	for(int i=0;i<=n;++i)
		b[i]=finv[n-i];
	fft(a,maxn<<1,1);
	fft(b,maxn<<1,1);
	for(int i=0;i<(maxn<<1);++i)
		c[i]=a[i]*b[i]%mod;
	fft(c,maxn<<1,-1);
	for(int i=1;i<=n;++i)
		printf("%lld\n",(n*fac[n]%mod*finv[i]%mod*finv[n-i]%mod-finv[i]*c[n+i]%mod+mod)%mod);
	return 0;
}