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 <iostream>
#include <cstdio>
#include <algorithm>
#define N 1000002
#define int long long
using namespace std;
const int mod=924844033;
const int G=5;
int head[N],ver[N*2],nxt[N*2],l;
int n,i,size[N],cnt[N],fac[N],inv[N],f[N],g[N],m=1,lim,r[N];
int read()
{
	char c=getchar();
	int w=0;
	while(c<'0'||c>'9') c=getchar();
	while(c<='9'&&c>='0'){
		w=w*10+c-'0';
		c=getchar();
	}
	return w;
}
void insert(int x,int y)
{
	l++;
	ver[l]=y;
	nxt[l]=head[x];
	head[x]=l;
}
void dfs(int x,int pre)
{
	size[x]=1;
	for(int i=head[x];i;i=nxt[i]){
		int y=ver[i];
		if(y!=pre){
			dfs(y,x);
			size[x]+=size[y];
			cnt[size[y]]++;
		}
	}
	cnt[n-size[x]]++;
}
int poww(int a,int b)
{
	int ans=1,base=a;
	while(b){
		if(b&1) ans=ans*base%mod;
		base=base*base%mod;
		b>>=1;
	}
	return ans;
}
void init()
{
	dfs(1,0);
	fac[0]=1;
	for(int i=1;i<=n;i++) fac[i]=fac[i-1]*i%mod;
	inv[n]=poww(fac[n],mod-2);
	for(int i=n-1;i>=0;i--) inv[i]=inv[i+1]*(i+1)%mod;
}
void NTT(int *a,int inv)
{
	for(int i=0;i<m;i++){
		if(i<r[i]) swap(a[i],a[r[i]]);
	}
	for(int l=2;l<=m;l<<=1){
		int mid=l/2;
		int cur=poww(G,(mod-1)/l);
		if(inv==-1) cur=poww(cur,mod-2);
		for(int i=0;i<m;i+=l){
			int omg=1;
			for(int j=0;j<mid;j++,omg=omg*cur%mod){
				int tmp=omg*a[i+j+mid]%mod;
				a[i+j+mid]=(a[i+j]-tmp+mod)%mod;
				a[i+j]=(a[i+j]+tmp)%mod;
			}
		}
	}
}
int C(int n,int m)
{
	return fac[n]*inv[m]%mod*inv[n-m]%mod;
}
signed main()
{
	n=read();
	for(i=1;i<n;i++){
		int u=read(),v=read();
		insert(u,v);
		insert(v,u);
	}
	init();
	for(i=1;i<=n;i++) f[i]=cnt[i]*fac[i]%mod;
	for(i=0;i<=n;i++) g[i]=inv[i];
	reverse(f,f+n+1);
	while(m<2*n+2) m<<=1;
	while((1<<lim)<m) lim++;
	for(i=0;i<m;i++) r[i]=(r[i>>1]>>1)|((i&1)<<(lim-1));
	NTT(f,1);NTT(g,1);
	for(i=0;i<m;i++) f[i]=f[i]*g[i]%mod;
	NTT(f,-1);
	int invm=poww(m,mod-2);
	for(i=0;i<m;i++) f[i]=f[i]*invm%mod;
	reverse(f,f+n+1);
	for(i=1;i<=n;i++) printf("%lld\n",(n*C(n,i)%mod-inv[i]*f[i]%mod+mod)%mod);
	return 0;
}
