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;
const int N=1000005,mod=924844033;
int n,fac[N],ifac[N],A[N],B[N],sz[N],r[N];
vector<int>g[N];
int pw(int a,int b){int r=1;for(;b;b>>=1,a=1ll*a*a%mod)if(b&1)r=1ll*r*a%mod;return r;}
int C(int a,int b){return 1ll*fac[a]*ifac[b]%mod*ifac[a-b]%mod;}
void ntt(int n,int *a,int f)
{
	int l=0;while((1<<l)<n)l++;
	for(int i=1;i<n;i++)r[i]=r[i>>1]>>1|((i&1)<<(l-1));
	for(int i=0;i<n;i++)if(i<r[i])swap(a[i],a[r[i]]);
	for(int i=2;i<=n;i<<=1)
	{
		int wn=pw(5,(mod-1)/i);
		if(f==-1)wn=pw(wn,mod-2);
		for(int j=0;j<n;j+=i)for(int k=0,w=1;k<(i>>1);k++,w=1ll*w*wn%mod)
		{
			int u=a[j+k],v=1ll*a[j+k+(i>>1)]*w%mod;
			a[j+k]=(u+v)%mod;a[j+k+(i>>1)]=(u-v+mod)%mod;
		}
	}
	if(f==-1)for(int i=0,iv=pw(n,mod-2);i<n;i++)a[i]=1ll*a[i]*iv%mod;
}
void dfs(int x,int p)
{
	sz[x]=1;
	for(int i=0;i<g[x].size();i++)if(g[x][i]!=p)dfs(g[x][i],x),sz[x]+=sz[g[x][i]];
	if(x!=1)A[sz[x]]++,A[n-sz[x]]++;
}
int main()
{
	scanf("%d",&n);
	for(int i=1,u,v;i<n;i++)
	{
		scanf("%d%d",&u,&v);
		g[u].push_back(v);g[v].push_back(u);
	}
	dfs(1,0);
	for(int i=fac[0]=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mod;
	ifac[n]=pw(fac[n],mod-2);
	for(int i=n;i;i--)ifac[i-1]=1ll*ifac[i]*i%mod;
	for(int i=0;i<=n;i++)A[i]=1ll*A[i]*fac[i]%mod;
	reverse(A,A+n+1);
	int nn=1;while(nn<=(n<<1))nn<<=1;
	for(int i=0;i<nn;i++)B[i]=ifac[i];
	ntt(nn,A,1);ntt(nn,B,1);
	for(int i=0;i<nn;i++)A[i]=1ll*A[i]*B[i]%mod;
	ntt(nn,A,-1);
	reverse(A,A+n+1);
	for(int i=0;i<=n;i++)A[i]=1ll*A[i]*ifac[i]%mod;
	for(int i=1;i<=n;i++)A[i]=(1ll*n*C(n,i)%mod-A[i]+mod)%mod;
	for(int i=1;i<=n;i++)printf("%d\n",A[i]);
	return 0;
} 