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<cstring>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
#define N 1000002
const int p=924844033;
struct edge{int to,nxxt;}e[N];
int head[N],cnt=1,n,siz[N],f[N],inv[N],finv[N],fa[N],ct[N],tt[N],rev[N];
int a[N],b[N],c[N];
inline void ins(int u,int v){e[cnt].to=v;e[cnt].nxxt=head[u];head[u]=cnt++;}
void df5(int te,int la)
{siz[te]=1;
	for(int i=head[te];i;i=e[i].nxxt)
	{
		int j=e[i].to;if(j==la)continue;
		fa[j]=te;df5(j,te);siz[te]+=siz[j];
	}
}
inline int C(int nn,int mm)
{
	if(nn<mm)return 0;
	return 1ll*f[nn]*finv[mm]%p*finv[nn-mm]%p;
}
inline int ksm(int d,int k){int ret=1;while(k){if(k&1)ret=1ll*ret*d%p;d=1ll*d*d%p;k>>=1;}return ret;}
inline void ntt(int x[],int len,int mde)
{
	for(int i=0;i<len;i++)if(i<rev[i])swap(x[i],x[rev[i]]);
	for(int i=2;i<=len;i<<=1)
	{
		int wn=ksm(5,(p-1)/i);if(mde<0)wn=ksm(wn,p-2);
		for(int j=0,stp=i>>1;j<len;j+=i)for(int k=j,w=1;k<j+stp;k++,w=1ll*w*wn%p)
		{
			int t1=x[k],t2=1ll*x[k+stp]*w%p;
			x[k]=(t1+t2)%p,x[k+stp]=(t1-t2+p)%p;
		}
	}if(mde<0)for(int i=0,te=ksm(len,p-2);i<len;i++)x[i]=1ll*x[i]*te%p;
}
int main()
{
    scanf("%d",&n);
	f[0]=finv[0]=f[1]=finv[1]=inv[1]=1;
	for(int i=2;i<=500000;i++)
	{
		inv[i]=1ll*(p-p/i)*inv[p%i]%p;
		f[i]=1ll*f[i-1]*i%p;finv[i]=1ll*finv[i-1]*inv[i]%p;
	}
	for(int i=1;i<n;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		ins(x,y),ins(y,x);
	}
	df5(1,1);
	for(int i=2;i<=n;i++)ct[siz[i]]++;
	for(int i=1;i<=n;i++)tt[n-i]=ct[i];
	for(int i=1;i<=n;i++)ct[i]+=tt[i];
	int len=1;while(len<(n<<1))len<<=1;
	for(int i=1;i<=len;i++){rev[i]=rev[i>>1]>>1;if(i&1)rev[i]|=len>>1;}
	for(int i=1;i<=n;i++)a[i]=1ll*f[i]*ct[i]%p;
	for(int i=0;i<=n;i++)b[i]=finv[n-i];
	ntt(a,len,1);ntt(b,len,1);
	for(int i=0;i<len;i++)a[i]=1ll*a[i]*b[i]%p;
	ntt(a,len,-1);
	for(int i=1;i<=n;i++)a[i]=a[i+n];
	for(int i=1;i<=n;i++)
	{
		int ts=(1ll*n*C(n,i)%p-1ll*finv[i]*a[i]%p+p)%p;
		printf("%d\n",ts);
	}
}
