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>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<50;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

const ll p=924844033,r=5;

ll Pow(ll n,ll m){
	ll res=1;
	while(m>0){
		if(m&1) (res*=n)%=p;
		(n*=n)%=p;
		m>>=1;
	}
	return res;
}

ll Inv(ll x){return Pow(x,p-2);}

void ntt(vl& a,bool B){
	int n=a.size(),h=0;
	while(1<<h<n) h++;
	for(int i=0;i<n;i++){
		int j=0;
		for(int k=0;k<h;k++) if(i&1<<k) j+=1<<(h-k-1);
		if(i<j) swap(a[i],a[j]);
	}
	for(int i=1;i<n;i*=2){
		ll w=Pow(r,(p-1)/(i*2));
		if(B) w=Inv(w);
		for(int j=0;j<n;j+=i*2){
			ll wn=1;
			for(int k=0;k<i;k++){
				ll s=a[j+k],t=a[i+j+k]*wn%p;
				a[j+k]=(s+t)%p;
				a[i+j+k]=(s-t+p)%p;
				(wn*=w)%=p;
			}
		}
	}
	if(B){
		ll v=Inv(n);
		for(int i=0;i<n;i++) (a[i]*=v)%=p;
	}
}

vl Conv(vl a,vl b){
	int s=a.size()+b.size()-1,t=1;
	while(t<s) t*=2;
	a.resize(t);
	b.resize(t);
	ntt(a,0);ntt(b,0);
	for(int i=0;i<t;i++) (a[i]*=b[i])%=p;
	ntt(a,1);
	a.resize(s);
	return a;
}

const int M=200005;
ll F[M];

void Init(){
	F[0]=1;
	for(int i=1;i<M;i++) F[i]=F[i-1]*i%p;
}

ll Div(ll n,ll m){
	return n*Pow(m,p-2)%p;
}

ll nCk(ll n,ll k){
	return Div(F[n],F[n-k]*F[k]%p);
}

int n;
vi s;
vvi g;

int dfs(int v,int w){
	s[v]++;
	for(auto u:g[v]) if(u!=w) s[v]+=dfs(u,v);
	return s[v];
}

int main(){
	cin>>n;
	s=vi(n);
	g=vvi(n);
	Init();
	for(int i=1;i<n;i++){
		int u,v;
		cin>>u>>v;
		u--;v--;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(0,-1);
	vl a(n+1),b(n+1);
	for(int i=0;i<n;i++){
		int S=n-1;
		for(auto v:g[i]) if(s[i]>s[v]) a[s[v]]++,S-=s[v];
		if(S) a[S]++;
	}
	for(int i=1;i<=n;i++){
		(a[i]*=F[i])%=p;
		b[i]=Inv(F[n-i]);
	}
	vl c=Conv(a,b);
	for(int i=1;i<=n;i++) cout<<(n*nCk(n,i)%p-Div(c[n+i],F[i])+p)%p<<endl;
}