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 clz(n) __builtin_clz(n)
typedef long long ll;
const int mod=924844033;
const int w_=5;
const int N=(1<<20)+1;
inline int ksm(ll a,int b,int c=1){for(;b;b/=2,a=a*a%mod)if(b&1)c=c*a%mod;return c;}
int jc[N],jc2[N];

void Init(){
	jc[0]=1;
	for(int i=1;i<N;i++)jc[i]=(ll)jc[i-1]*i%mod;
	jc2[N-1]=ksm(jc[N-1],mod-2);
	for(int i=N-1;i;i--)jc2[i-1]=(ll)jc2[i]*i%mod;
}

int n_,lgn_;
int rev_[N],omg_[N];
void FFT_init(int n){
    lgn_=32-clz(n-1);n_=1<<lgn_;
    rev_[0]=0;omg_[0]=1;
    ll d=ksm(w_,441<<(21-lgn_));
    for(int i=1;i<n_;i++){
        omg_[i]=omg_[i-1]*d%mod;
        rev_[i]=rev_[i/2]>>1|(i&1)<<(lgn_-1);
    }
}

ll ww[N];
void FFT(int*a,int fg=0){
    for(int i=0;i<n_;i++)if(rev_[i]<i)swap(a[i],a[rev_[i]]);
    int s,t;
    for(int i=0;i<lgn_;i++){
        t=1<<i;
        for(int j=0;j<1<<i;++j)ww[j]=omg_[j<<(lgn_-i-1)];
        for(int j=0;j<n_;j+=t<<1){
            int*at=a+j,*bt=a+j+t;
            for(int k=0;k<t;++k){
                s=ww[k]*bt[k]%mod;
                bt[k]=at[k]-s;bt[k]+=bt[k]>>31&mod;
				at[k]+=s-mod;at[k]+=at[k]>>31&mod;
            }
        }
    }
    if(fg){
        reverse(a+1,a+n_);
        ll iv=ksm(n_,mod-2);
        for(int i=0;i<n_;++i)a[i]=a[i]*iv%mod;
    }
}

int C(int n,int k){
	if(k<0)return 0;
	if(n<0){
		int res=C(k-n-1,k);
		if(k&1)return (mod-res)%mod;
		else return res;
	}
	if(n<k)return 0;
	return (ll)jc[n]*jc2[k]%mod*jc2[n-k]%mod;
}
int n,f[N],a[N],sz[N],fa[N];
vector<int>g[N];
void dfs(int x){
	sz[x]=1;
	for(int i:g[x])if(i!=fa[x]){
		fa[i]=x;dfs(i);
		f[sz[i]]++;sz[x]+=sz[i];
	}
	f[n-sz[x]]++;
}


int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;Init();
	for(int i=1;i<n;++i){
		int x,y;cin>>x>>y;
		g[x].push_back(y);g[y].push_back(x);
	}
	dfs(1);
	for(int i=0;i<=n;++i)f[i]=(ll)f[i]*jc[i]%mod;
	for(int i=0;i<=n;++i)a[i]=jc2[i];
	reverse(a,a+n+1);
	FFT_init(2*n+1);
	FFT(f);FFT(a);
	for(int i=0;i<n_;++i)f[i]=(ll)f[i]*a[i]%mod;
	FFT(f,1);
	for(int i=1;i<=n;++i){
		int ret=(ll)C(n,i)*n%mod,res=(ll)f[i+n]*jc2[i]%mod;
		ret=(ret-res+mod)%mod;
		cout<<ret<<'\n';
	}
	return 0;
}