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 <algorithm>
#define IL inline
#define N 200010
#define S 524291
#define REP(a,b,c) for(a=b;a<=c;a++)
#define PER(a,b,c) for(a=b;a>=c;a--)
using namespace std;
const int mod=924844033;
int head[N],lt[N<<1],nt[N<<1],ap=1,q[N],fa[N],sz[N];
int n,L,R[S],G[S],F[S],inv[N];
IL int rd(){
	int res=0;char c;while((c=getchar())<'0'||c>'9');
	while(c>='0'&&c<='9')res=res*10+c-'0',c=getchar();return res;
}
IL int qpow(int a,int b){
	int res=1;while(b){if(b&1)res=1LL*res*a%mod;a=1LL*a*a%mod,b>>=1;}return res;
}
IL void link(int a,int b){
	lt[++ap]=b,nt[ap]=head[a],head[a]=ap;
	lt[++ap]=a,nt[ap]=head[b],head[b]=ap;
}
IL void bfs(){
	int fr=1,tp=1,u,x,v;q[1]=1;
	while(fr<=tp){
		for(x=head[u=q[fr++]];x;x=nt[x])
			if((v=lt[x])!=fa[u])q[++tp]=v,fa[v]=u;
	}
	for(fr--;fr;fr--,G[n-sz[u]]--)
		for(x=head[u=q[fr]],sz[u]++;x;x=nt[x])
			if((v=lt[x])!=fa[u])sz[u]+=sz[v],G[sz[v]]--;
	G[n]+=n;
	REP(u,0,n)if(G[u]<0)G[u]+=mod;
}
IL void NTT(int *a,int f){
	int i,j,k,gn,g,x,y;
	REP(i,0,n)if(R[i]>i)swap(a[i],a[R[i]]);
	for(i=1;i<n;i<<=1){
		gn=qpow(5,(mod-1)/(i<<1));
		for(j=0;j<n;j+=i<<1){
			g=1;
			for(k=0;k<i;k++,g=1LL*g*gn%mod){
				x=a[j+k],y=1LL*g*a[j+k+i]%mod;
				a[j+k]=(x+y)%mod,a[j+k+i]=(x-y+mod)%mod;
			}
		}
	}
	if(f==-1){
		reverse(a+1,a+n);g=qpow(n,mod-2);
		REP(i,0,n-1)a[i]=1LL*a[i]*g%mod;
	}
}
int main(){
  n=rd();int i,j=1;REP(i,2,n)link(rd(),rd());bfs();
	REP(i,1,n)G[i]=1LL*G[i]*(j=1LL*j*i%mod)%mod;
	inv[n]=j=qpow(j,mod-2);PER(i,n,1)inv[i-1]=F[n-i+1]=j=1LL*j*i%mod;
	j=n;
	for(n=1;n<=j<<1;n<<=1)L++;
	REP(i,0,n-1)R[i]=(R[i>>1]>>1|(i&1)<<(L-1));
	NTT(F,1),NTT(G,1);
	REP(i,0,n-1)F[i]=1LL*F[i]*G[i]%mod;
	NTT(F,-1);
	REP(i,1,j)printf("%lld\n",1LL*F[j+i]*inv[i]%mod);
  return 0;
}