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 fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
typedef long long ll;
using namespace std;
inline int read(){
	int n=0,f=1;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if (c=='-') c=getchar(),f=-1;
	for(;c>='0'&&c<='9';c=getchar()) n=n*10+c-48;
	return n*f;
}
const int maxn=1<<20,mo=924844033;
int i,j,n,k[maxn],re,ne[maxn],g[maxn],num;
int s[maxn],a[maxn],f[maxn],len,t;
int ji[maxn],ni[maxn],ans[maxn],b[maxn],w[maxn];
void add(int x,int y){
	ne[++num]=k[x],k[x]=num,g[num]=y;
}
void dfs(int x,int y){
	int i;
	s[x]=1;
	for(i=k[x];i;i=ne[i]){
		int go=g[i];
		if (go!=y)dfs(go,x),s[x]+=s[go];
	}
}
int qs(int x,int y){
	int s=1;
	for(;y;y>>=1,x=(ll)x*x%mo) if (y&1) s=(ll)s*x%mo;
	return s;
}
void mod(int &x){
	x-=(x>=mo)?mo:0;
}
void dft(int *a,bool sig){
	int i,j,k,m=1,half,ws,o;
	fo(i,0,len-1) g[f[i]]=a[i];
	for(;m<len;){
		half=m,m<<=1,ws=sig?w[len/m]:w[len-len/m];
		for(i=0;i<len;i+=m){
			for(j=i,o=1;j<i+half;j++,o=(ll)o*ws%mo){
				int u=g[j],v=(ll)g[j+half]*o%mo;
				mod(g[j]+=v),mod(g[j+half]=u+mo-v);
			}
		}
	}
	if (sig){
		fo(i,0,len-1) a[i]=g[i];
	}else fo(i,0,len-1) a[i]=(ll)g[i]*re%mo;
}
int main(){
	n=read();
	fo(i,1,n-1) {
		int x=read(),y=read();
		add(x,y),add(y,x);
	}
	ji[0]=1;
	fo(i,1,n) ji[i]=(ll)ji[i-1]*i%mo;
	ni[n]=qs(ji[n],mo-2);
	fd(i,n,1) ni[i-1]=(ll)ni[i]*i%mo;
	dfs(1,0);
	a[n]=n;
	fo(i,2,n) a[s[i]]--,a[n-s[i]]--;
	fo(i,1,n) a[i]=a[i]<0?a[i]+mo:a[i];
	b[0]=1;
	fo(i,1,n) a[i]=(ll)a[i]*ji[i]%mo,b[i]=ni[i];
	fo(i,1,n>>1) swap(a[i],a[n-i+1]);
	len=1;
	for(;len<=n*2;len<<=1,t++);
	w[0]=1,w[1]=qs(5,(mo-1)/len);
	fo(i,2,len) w[i]=(ll)w[i-1]*w[1]%mo;
	re=qs(len,mo-2);
	fo(i,0,len-1){
		int x=i,y=t-1;
		for(;x;x>>=1,y--) f[i]+=(x&1)<<y;
	}
	dft(a,1),dft(b,1);
	fo(i,0,len-1) a[i]=(ll)a[i]*b[i]%mo;
	dft(a,0);
	fo(i,1,n) ans[i]=a[i];
	fo(i,1,n>>1) swap(ans[i],ans[n-i+1]);
	fo(i,1,n) printf("%d\n",(ll)ans[i]*ni[i]%mo);
}