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 ll long long
using namespace std;
const int N=600010;
const int mod=924844033;
struct node{
	int v,nxt;
}e[N<<1];
int n,cnt=0,head[N],siz[N],cn[N],jc[N],inv[N];
int a[N],b[N],lim=1,l=0,r[N];
void add(int u,int v){
	e[++cnt].v=v;
	e[cnt].nxt=head[u];
	head[u]=cnt;
}
ll poww(ll x,int y){
	ll ans=1;
	while(y){
		if(y&1){
			ans=ans*x%mod;
		}
		x=x*x%mod;
		y>>=1;
	}
	return ans;
}
void dfs(int s,int fa){
	siz[s]=1;
	for(int i=head[s];i;i=e[i].nxt){
		int v=e[i].v;
		if(v==fa) continue;
		dfs(v,s);
		siz[s]+=siz[v];
		++cn[siz[v]];
	}
	++cn[n-siz[s]];
}
int C(int n,int m){
	return 1ll*jc[n]*inv[m]%mod*inv[n-m]%mod;
}
void ntt(int *a,int f){
	for(int i=0;i<lim;i++){
		if(i<r[i]) swap(a[i],a[r[i]]);
	}
	for(int i=1;i<lim;i<<=1){
		int wn=poww(f==1?5:554906420,(mod-1)/(i<<1));
		for(int j=0;j<lim;j+=(i<<1)){
			ll w=1;
			for(int k=0;k<i;k++,w=1ll*w*wn%mod){
				ll x=a[j+k],y=1ll*w*a[j+i+k]%mod;
				a[j+k]=(x+y)%mod;
				a[j+i+k]=(x-y+mod)%mod;
			}
		}
	}
	if(f==-1){
		ll inv=poww(lim,mod-2);
		for(int i=0;i<lim;i++){
			a[i]=1ll*a[i]*inv%mod;
		} 
	}
}
int main(){
//	freopen("k.in","r",stdin);
//	freopen("k.out","w",stdout);
	scanf("%d",&n);
	for(int i=1,x,y;i<n;i++){
		scanf("%d%d",&x,&y);
		add(x,y);add(y,x);
	}
	dfs(1,0);
	jc[0]=1;
	for(int i=1;i<=n;i++){
		jc[i]=1ll*jc[i-1]*i%mod;
	}
	inv[n]=poww(jc[n],mod-2);
	for(int i=n-1;i>=0;i--){
		inv[i]=1ll*inv[i+1]*(i+1)%mod;
	}
	for(int i=0;i<=n;i++){
		a[i]=1ll*cn[i]*jc[i]%mod;
		b[i]=inv[i];
	}
	reverse(a,a+n+1);
	while(lim<=(n<<1)){
		lim<<=1;
		l++;
	}
	for(int i=0;i<lim;i++){
		r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
	} 
	ntt(a,1);
	ntt(b,1);
	for(int i=0;i<lim;i++){
		a[i]=1ll*a[i]*b[i]%mod;
	}
	ntt(a,-1);
	reverse(a,a+n+1);
	for(int i=1;i<=n;i++){
		printf("%lld\n",(1ll*n*C(n,i)%mod-1ll*inv[i]*a[i]%mod+mod)%mod);
	}
}