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
#include<bits/stdc++.h>
#define pi 3.141592653589793238462643387
#define mo 924844033ll
#define ll long long
#define N 600010
#define int long long
using namespace std;
ll n,vv=1,rev[N],len,x,y;
ll f[N],g[N],h[N],ec,v[N],nxt[N],sz[N],c[N],jc[N],ij[N];
void add(int x,int y){v[++ec]=y;nxt[ec]=h[x];h[x]=ec;}
ll qp(ll x,ll y){
    if(!y)return 1;
    if(y==1)return x%mo;
    ll r=qp(x,y/2);
    if(y&1)return r*r%mo*x%mo;
    return r*r%mo;
}
void fft(ll *a,ll t){
    for(ll i=0;i<vv;i++)
		if(i<rev[i])
			swap(a[i],a[rev[i]]);
    for(ll i=1;i<vv;i*=2){
        ll w;
        if(t==1)w=qp(5,(mo-1)/(i*2));
        else w=qp(554906420,(mo-1)/(i*2));
        for(ll r=i*2,j=0;j<vv;j+=r){
            ll tp=1;
            for(ll k=0;k<i;k++,tp=(tp*w)%mo){
                ll tx=a[j+k],ty=(a[j+i+k]*tp)%mo;
                a[j+k]=(tx+ty)%mo;
				a[j+i+k]=(tx-ty+mo)%mo;
            }
        }
    }
    if(t==1)return;
    ll iv=qp(vv,mo-2);
    for(int i=0;i<vv;i++)
    	a[i]=a[i]*iv%mo;
}
void dfs(int x,int fa){
	sz[x]=1;
	for(int i=h[x];i;i=nxt[i])
		if(v[i]!=fa){
			dfs(v[i],x);
			sz[x]+=sz[v[i]];
			c[sz[v[i]]]++;
		}
	c[n-sz[x]]++;
}
void init(){
	jc[0]=1;
	for(int i=1;i<N;i++)
		jc[i]=jc[i-1]*i%mo;
	ij[N-1]=qp(jc[N-1],mo-2);
	for(int i=N-1;i;i--)
		ij[i-1]=ij[i]*i%mo;
}
int cc(int x,int y){
	return jc[x]*ij[y]%mo*ij[x-y]%mo;
}
signed main(){
    cin>>n;
    for(int i=1;i<n;i++){
    	int a,b;
    	cin>>a>>b;
    	add(a,b);
    	add(b,a);
	}
	dfs(1,0);
	init();
	for(int i=1;i<=n;i++)
		f[i]=c[i]*jc[i]%mo;
	reverse(f,f+n+1);
	for(int i=0;i<=n;i++)
		g[i]=ij[i];
	vv=1;
	int len=0;
	while(vv<=2*n)vv*=2,len++;
    for(ll i=0;i<vv;i++)
        rev[i]=(rev[i>>1]>>1)|((i&1)<<(len-1));
   	fft(f,1);
    fft(g,1);
	for(int i=0;i<vv;i++)
		f[i]=f[i]*g[i]%mo;
	fft(f,-1);
	reverse(f,f+n+1);
	for(int i=1;i<=n;i++)
		cout<<(n*cc(n,i)%mo-ij[i]*f[i]%mo+mo)%mo<<'\n';
}