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
//ΔAGC005F
#include<iostream>
#include<cstdio>
#include<fstream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<bitset>
#include<cmath>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;
typedef double DB;
const int N = 1<<19;
const int MO = 924844033;
int ad(int x,int y){
	x+=y;
	if(x>=MO)
		x-=MO;
	if(x<0)
		x+=MO;
	return x;
}
int mul(int x,int y){
	return (LL)x*y%MO;
}
int fpow(int x,int y=MO-2){
	if(!y)
		return 1;
	int z=fpow(x,y>>1);
	z=mul(z,z);
	if(y&1)
		return mul(z,x);
	return z;
}
void DFT(int C[],int n,int fl){
	int i,j,k,l;
	int u,v,w,cur;
	i=n>>1;
	for(j=1;j<n;j=j+1){
		if(i<j)
			swap(C[i],C[j]);
		for(k=n>>1;i&k;k>>=1)
			i^=k;
		i^=k;
	}
	for(l=2;l<=n;l<<=1){
		w=fpow(5,(MO-1)/l);
		if(fl==-1)
			w=fpow(w);
		for(i=0;i<n;i=i+l){
			cur=1;
			for(j=i;j<i+(l>>1);j=j+1){
				u=C[j],v=mul(C[j+(l>>1)],cur);
				C[j]=ad(u,v),C[j+(l>>1)]=ad(u,-v);
				cur=mul(cur,w);
			}
		}
	}
}
int m,A[N],B[N];
void FFT(){
	int i,l,o;
	for(l=1;l<m+m;l<<=1);
	DFT(A,l,1);
	DFT(B,l,1);
	for(i=0;i<l;i=i+1)
		A[i]=mul(A[i],B[i]);
	DFT(A,l,-1);
	o=fpow(l);
	for(i=0;i<l;i=i+1)
		A[i]=mul(A[i],o);
}
int fac[N],ifac[N];
int n,t[N],s[N];
vector<int> v[N];
void dfs(int u,int fa){
	int i=v[u].size(),x;
	t[u]=1;
	while(i--){
		x=v[u][i];
		if(x!=fa)
			dfs(x,u);
	}
	t[fa]+=t[u];
	if(t[u]<n){
		s[t[u]]++;
		s[n-t[u]]++;
	}
}
int main()
{
	int i,x,y;
	scanf("%d",&n);
	fac[0]=1;
	for(i=1;i<=n;i=i+1)
		fac[i]=mul(fac[i-1],i);
	for(i=0;i<=n;i=i+1)
		ifac[i]=fpow(fac[i]);
	for(i=1;i<n;i=i+1){
		scanf("%d%d",&x,&y);
		v[x].push_back(y);
		v[y].push_back(x);
	}
	dfs(1,0);
	m=n+1;
	for(i=1;i<=n;i=i+1)
		A[i]=mul(s[i],fac[i]);
	for(i=0;i<=n;i=i+1)
		B[i]=ifac[n-i];
	FFT();
	for(i=1;i<=n;i=i+1){
		x=mul(mul(n,fac[n]),mul(ifac[i],ifac[n-i]));
		y=mul(A[n+i],ifac[i]);
		printf("%d\n",ad(x,-y));
	}
	return 0;
}