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 maxn=2e5+10;
const int mod=924844033;
const int G=5;
const int invG=554906420;
int n,cnt_edge;
int head[maxn],size[maxn],cnt[maxn];
struct edge{int to,nxt;}e[maxn<<1]; 
inline int read()
{
    char c=getchar();int res=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9')res=res*10+c-'0',c=getchar();
    return res*f;
}
inline void add_edge(int u,int v)
{
	e[++cnt_edge]=(edge){v,head[u]};
	head[u]=cnt_edge;
}
void dfs(int x,int fa)
{
	size[x]=1;
	for(int i=head[x];i;i=e[i].nxt)
	{
		int y=e[i].to;
		if(y==fa)continue;
		dfs(y,x);size[x]+=size[y];
		cnt[size[y]]++;
	}
	cnt[n-size[x]]++;
}
int fac[maxn],invfac[maxn];
inline int power(int x,int k)
{
	int res=1;
	while(k)
	{
		if(k&1)res=1ll*res*x%mod;
		x=1ll*x*x%mod;k>>=1;
	}
	return res;
}
inline int C(int n,int m){return 1ll*fac[n]*invfac[n-m]%mod*invfac[m]%mod;}
int pos[maxn<<2],f[maxn<<2],g[maxn<<2];
inline void NTT(int* a,int len,int op)
{
    int lim=1,cnt=0;
    while(lim<len)lim<<=1,cnt++;
    for(int i=0;i<lim;i++)pos[i]=(pos[i>>1]>>1)|((i&1)<<(cnt-1));
    for(int i=0;i<lim;i++)if(i<pos[i])swap(a[i],a[pos[i]]);
    for(int l=1;l<lim;l<<=1)
    {
        int wn=power(op==1?G:invG,(mod-1)/(l<<1));
        for(int i=0;i<lim;i+=l<<1)
        {
            int w=1;
            for(int j=0;j<l;j++,w=1ll*w*wn%mod)
            {
                int x=a[i+j],y=1ll*w*a[i+l+j]%mod;
                a[i+j]=(x+y)%mod;a[i+l+j]=(x-y+mod)%mod;
            }
        }
    }
    if(op==1)return;
    int inv=power(lim,mod-2);
    for(int i=0;i<lim;i++)a[i]=1ll*a[i]*inv%mod;
}
int main()
{
	n=read();
	for(int i=1;i<n;i++)
	{
		int u=read(),v=read();
		add_edge(u,v),add_edge(v,u);
	}
	dfs(1,0);
	fac[0]=1;for(int i=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mod;
	invfac[n]=power(fac[n],mod-2);
	for(int i=n;i;i--)invfac[i-1]=1ll*invfac[i]*i%mod;
	for(int i=1;i<=n;i++)f[i]=1ll*cnt[i]*fac[i]%mod;
	reverse(f,f+n+1);
	for(int i=0;i<=n;i++)g[i]=invfac[i];
	int lim=1;
	while(lim<(n<<1))lim<<=1;
	NTT(f,lim,1),NTT(g,lim,1);
	for(int i=0;i<lim;i++)f[i]=1ll*f[i]*g[i]%mod;
	NTT(f,lim,-1);
	reverse(f,f+n+1);
	for(int i=1;i<=n;i++)printf("%d\n",(1ll*n*C(n,i)-1ll*invfac[i]*f[i]%mod+mod)%mod);
	return 0;
}