Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<algorithm>
using namespace std;

const int N=200005,P=924844033,G=5;
int n,tot,nxt[2*N],fst[N],to[2*N],size[N],f[N],rf[N],c[N*4],d[N*4];

int pw(int a,int x)
{
	int ans=1;
	for(;x;x>>=1,a=1ll*a*a%P)
		if(x&1)ans=1ll*ans*a%P;
	return ans;
}

void dfs(int u,int fa)
{
	size[u]=1;
	for(int i=fst[u];i;i=nxt[i])
		if(to[i]!=fa)
		{
			dfs(to[i],u);
			size[u]+=size[to[i]];
		}
	if(fa)d[size[u]]++,d[n-size[u]]++;
}

void ntt(int *a,int n,int inv)
{
	for(int i=0,j=0;i<n;i++)
	{
		if(i>j)swap(a[i],a[j]);
		for(int k=n>>1;(j^=k)<k;k>>=1);
	}
	for(int i=2;i<=n;i<<=1)
	{
		int m=i>>1,wn=pw(G,(P-1)+inv*(P-1)/i);
		for(int *p=a;p<a+n;p+=i)
		{
			int w=1;
			for(int j=0;j<m;j++)
			{
				int t=1ll*w*p[m+j]%P;
				p[m+j]=(p[j]-t)%P;
				p[j]=(p[j]+t)%P;
				w=1ll*w*wn%P;
			}
		}
	}
	if(inv==-1)
	{
		int t=pw(n,P-2);
		for(int i=0;i<n;i++)a[i]=1ll*a[i]*t%P;
	}
}

int main()
{
	scanf("%d",&n);
	for(int i=1,u,v;i<n;i++)
	{
		scanf("%d%d",&u,&v);
		nxt[++tot]=fst[u];fst[u]=tot;to[tot]=v;
		nxt[++tot]=fst[v];fst[v]=tot;to[tot]=u;
	}
	dfs(1,0);
	for(int i=1;i<n;i++)d[i]*=-1;
	d[n]=n;
	for(int i=f[0]=1;i<=n;i++)f[i]=1ll*f[i-1]*i%P;
	rf[n]=pw(f[n],P-2);
	for(int i=n;i;i--)rf[i-1]=1ll*rf[i]*i%P;
	for(int i=0;i<=n;i++)d[i]=1ll*d[i]*f[i]%P;
	for(int i=0;i<=n/2;i++)swap(d[i],d[n-i]);
	for(int i=0;i<=n;i++)c[i]=rf[i];
	int l;
	for(l=1;l<=n+n;l<<=1);
	ntt(c,l,1),ntt(d,l,1);
	for(int i=0;i<l;i++)c[i]=1ll*c[i]*d[i]%P;
	ntt(c,l,-1);
	for(int i=1;i<=n;i++)
	{
		int ans=1ll*c[n-i]*rf[i]%P;
		printf("%d\n",(ans+P)%P);
	}
	return 0;
}