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>
using namespace std;
#define int long long
const int N=1300005,M=924844033;
int ne[N],tot,fi[N],zz[N],size[N],a[N],b[N],fac[N],inv[N],n,m,x,y;
void jb(int x,int y){
	ne[++tot]=fi[x];
	fi[x]=tot;
	zz[tot]=y;
}
void dfs(int x,int y){
	size[x]=1;
	for (int i=fi[x];i;i=ne[i])
		if (zz[i]!=y)dfs(zz[i],x),size[x]+=size[zz[i]];
}
int ksm(int x,int y){
	int ans=1;
	for (;y;y>>=1,x=x*x%M)
		if (y&1)(ans*=x)%=M;
	return ans;	
}
int C(int x,int y){
	if (x<y)return 0;
	return fac[x]*inv[y]%M*inv[x-y]%M;
}
void init(int y[],int len){
	for (int i=1,j=len/2;i<len-1;i++){
		if (i<j)swap(y[i],y[j]);
		int k=len/2;
		while (j>=k){
			j-=k;
			k/=2;
		}
		j+=k;
	}
}
void ntt(int y[],int len,int opt){
	init(y,len);
	for (int h=2;h<=len;h<<=1){
		int wn=ksm(5,(M-1)/h);
		if (opt==-1)wn=ksm(wn,M-2);
		for (int j=0;j<len;j+=h){
			int w=1;
			for (int k=j;k<j+h/2;k++){
				int u=y[k],v=w*y[k+h/2]%M;
				y[k]=(u+v)%M;
				y[k+h/2]=(u-v+M)%M;
				w=w*wn%M;
			}
		}
	}
	if (opt==1)return;
	int temp=ksm(len,M-2);
	for (int i=0;i<len;i++)(y[i]*=temp)%=M;
}
signed main(){
	scanf("%lld",&n);
	fac[0]=inv[0]=1;
	for (int i=1;i<N;i++)fac[i]=fac[i-1]*i%M,inv[i]=ksm(fac[i],M-2);
	for (int i=1;i<n;i++){
		scanf("%lld%lld",&x,&y);
		jb(x,y);jb(y,x);
	}
	dfs(1,0);
	for (int i=0;i<=n;i++)b[i]=inv[i];
	for (int i=2;i<=n;i++)a[size[i]]++,a[n-size[i]]++;
	for (int i=0;i<=n;i++)(a[i]*=fac[i])%=M;
	reverse(a,a+n+1);
	int len=1;
	while (len<n*2+5)len*=2;
	ntt(a,len,1);ntt(b,len,1);
	for (int i=0;i<len;i++)(a[i]*=b[i])%=M;
	ntt(a,len,-1);
	for (int i=n+1;i<len;i++)a[i]=0;
	reverse(a,a+n+1);
	for (int i=1;i<=n;i++)printf("%lld\n",(n*C(n,i)-a[i]*inv[i]%M+M)%M);
	return 0;
}