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<cstdio>
#include<cstring>
#include<algorithm>
#define gt getchar()
#define ll long long
#define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
inline int in()
{
	int k=0;char ch=gt;
	while(ch<'-')ch=gt;
	while(ch>'-')k=k*10+ch-'0',ch=gt;
	return k;
}
const int YL=924844033,N=2e6+5;
int rev[N];
inline int ksm(int a,int k){int r=1;while(k){if(k&1)r=1ll*r*a%YL;a=1ll*a*a%YL,k>>=1;}return r;}
inline int MO(const int &a){return a>=YL?a-YL:a;}
#define Wn (ksm(5,opt==1?(YL-1)/st:YL-1-(YL-1)/st))

void ntt(int *a,int len,int opt=1)
{
    for(int i=0;i<len;++i)if(i<rev[i])std::swap(a[i],a[rev[i]]);
    for(int st=2,m=1;st<=len;st<<=1,m<<=1)
        for(int *p=a,x,y,wn=Wn;p!=a+len;p+=st)
            for(int k=0,wnk=1;k<m;++k,wnk=1ll*wnk*wn%YL)
                x=p[k],y=1ll*wnk*p[k+m]%YL,p[k]=MO(x+y),p[k+m]=MO(x-y+YL);
    if(opt==1)return;int inv=ksm(len,YL-2);
    for(int i=0;i<len;++i)a[i]=1ll*a[i]*inv%YL;
}

int fac[N],fnv[N],a[N],b[N],num[N],head[N],to[N<<1],nxt[N<<1],cnt,siz[N],n;
inline int C(int x,int y){return y>x?0:1ll*fac[x]*fnv[y]%YL*fnv[x-y]%YL;}
inline void add(int u,int v)
{
	to[++cnt]=v,nxt[cnt]=head[u],head[u]=cnt;
	to[++cnt]=u,nxt[cnt]=head[v],head[v]=cnt;
}
void dfs(int u,int pa=0)
{
	siz[u]=1;
	for(int i=head[u],v;i;i=nxt[i])
		if((v=to[i])!=pa)
			dfs(v,u),++num[siz[v]],++num[n-siz[v]],siz[u]+=siz[v];
}
int main()
{
	n=in();fac[0]=1;
	for(int i=2;i<=n;++i)add(in(),in());dfs(1);
	for(int i=1;i<=n;++i)fac[i]=1ll*fac[i-1]*i%YL;fnv[n]=ksm(fac[n],YL-2);
	for(int i=n;i>=1;--i)fnv[i-1]=1ll*fnv[i]*i%YL;
	for(int i=0;i<=n;++i)a[i]=1ll*num[i]*fac[i]%YL;
	for(int i=0;i<=n;++i)b[i]=fnv[n-i];int len=1,qwq=0;
	while(len<(n<<1))len<<=1,++qwq;--qwq;
	for(int i=0;i<len;++i)rev[i]=(rev[i>>1]>>1)|((i&1)<<qwq);
	ntt(a,len),ntt(b,len);
	for(int i=0;i<len;++i)a[i]=1ll*a[i]*b[i]%YL;
	ntt(a,len,-1);
	for(int i=1;i<=n;++i)
		printf("%d\n",MO(1ll*n*C(n,i)%YL-1ll*fnv[i]*a[n+i]%YL+YL));
	return 0;
}
