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;
#define nn 65536*8
#define lln (16+3+1-1)
#define lf double 
#define ll long long
#define mod 924844033
#define g 5
ll qpow(ll x,ll y)
{
	if(y<0) y+=mod-1;
	ll res=1;
	while(y)
	{
		if(y&1) res=res*x%mod;
		x=x*x%mod;y=y/2;
	}
	return res;
}
ll pos[nn*2];
void ntt(ll a[],ll cid)
{
	for(ll i=0;i<nn*2;i++) if(pos[i]>i)swap(a[pos[i]],a[i]);
	for(ll p=1;p<nn*2;p<<=1)
	{
		ll m=p+p;
		ll w=qpow(g,(mod-1)/m*cid);
		for(ll i=0;i<nn*2;i+=m)
		{
			ll wx=1;
			for(ll j=0;j<p;j++,wx=wx*w%mod)
			{
				ll c1=a[i+j],c2=a[i+j+p]*wx%mod;
				a[i+j]=(c1+c2)%mod,a[i+j+p]=(c1-c2)%mod;
			}
		}
	}
}
ll n;
vector<ll> v[nn*2];ll siz[nn*2];ll a[nn*2];
void dfs(ll x,ll p)
{
	siz[x]=1;
	for(ll i=0;i<(ll)v[x].size();i++) if(v[x][i]!=p) dfs(v[x][i],x),siz[x]+=siz[v[x][i]],a[siz[v[x][i]]]++;
	a[n-siz[x]]++;
}
ll A[nn*3],B[nn*3],C[nn*3];
ll f[nn*2],rf[nn*2];
ll c(ll x,ll y)
{
	if(x>y) return 0;
	return f[y]*rf[x]%mod*rf[y-x]%mod;
}
int main()
{
	f[0]=1;for(ll i=1;i<nn;i++) f[i]=f[i-1]*i%mod;
	rf[nn-1]=qpow(f[nn-1],mod-2);for(ll i=nn-2;i>=0;i--) rf[i]=rf[i+1]*(i+1)%mod;
	scanf("%lld",&n);
	for(ll i=1;i<n;i++)
	{
		ll x,y;scanf("%lld%lld",&x,&y);
		v[x].push_back(y);v[y].push_back(x);
	}
	dfs(1,0);
	
	for(ll i=0;i<nn/2;i++) A[i+nn/2]=a[i]*f[i]%mod;
	for(ll i=0;i<nn/2;i++) B[nn/2-i]=rf[i];//cout<<B[0];
//	for(ll i=0;i<8;i++) cout<<A[i+nn/2+1]<<"*"<<B[nn/2-i]<<"\n";
	for(ll i=0;i<nn*2;i++)
		pos[i]=(((pos[i>>1])>>1)+(((i&1)<<lln)));
	ntt(A,1);ntt(B,1);
	for(ll i=0;i<nn*2;i++) C[i]=A[i]*B[i]%mod;
	ntt(C,-1);
	ll r=qpow(nn*2,mod-2);
	for(ll i=0;i<nn*2;i++) C[i]=C[i]*r%mod;
	//	cout<<C[nn/2*2]<<" ";


	for(ll i=1;i<=n;i++) printf("%lld\n",((n*c(i,n)%mod-C[nn/2*2+i]*rf[i]%mod)%mod+mod)%mod);
}