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>
using namespace std;
const int maxn=600005,maxe=maxn<<1,TT=924844033;
int n,L,len,sz[maxn],rev[maxn];
int e,lnk[maxn],nxt[maxe],son[maxe];
long long fct[maxn],ifct[maxn],G[2][maxn],a[maxn],b[maxn],c[maxn];
inline char nc(){
	static char buf[100000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int read(){
	int ret=0;bool f=0;char ch=nc();
	while(ch>'9'||ch<'0') f^=ch=='-',ch=nc();
	while(ch<='9'&&ch>='0') ret=ret*10+ch-'0',ch=nc();
	return f?-ret:ret;
}
inline void add_e(int x,int y){son[++e]=y;nxt[e]=lnk[x];lnk[x]=e;}
void DFS(int x,int fa){
	sz[x]=1;
	for (int i=lnk[x];i;i=nxt[i]) if(son[i]^fa)
	DFS(son[i],x),sz[x]+=sz[son[i]],a[sz[son[i]]]--;
	if(~fa) a[n-sz[x]]--;a[n]++;
}
inline long long pw(long long x,long long y){
	long long ret=1,w=x%TT;
	while(1){
		if(y&1) ret=ret*w%TT;
		if(!(y>>=1)) return ret;
		w=w*w%TT;
	}
}
inline void rsv(long long*A){for (int i=0;i<n-i;i++) swap(A[i],A[n-i]);}
inline void NTT(long long*A,bool flg){
	for (int i=0;i<len;i++) if(i<rev[i]) swap(A[i],A[rev[i]]);
	for (int i=1;(i<<1)<=len;i<<=1)
	for (int j=0;j<len;j+=(i<<1))
	for (int k=0;k<i;k++){
		long long x=A[j|k],y=A[i|j|k]*G[flg][i|k]%TT;
		A[i|j|k]=x<y?x-y+TT:x-y;A[j|k]=x+y>=TT?x+y-TT:x+y;
	}
	if(flg){
		int inv=pw(len,TT-2);
		for (int i=0;i<len;i++) A[i]=A[i]*inv%TT;
	}
}
int main(){
	n=read();
	for (int i=fct[0]=1;i<=n;i++) fct[i]=fct[i-1]*i%TT;
	ifct[n]=pw(fct[n],TT-2);
	for (int i=n;i;i--) ifct[i-1]=ifct[i]*i%TT;
	for (L=0,len=1;len<=(n<<1);L++,len<<=1);
	for (int i=1;i<len;i++) rev[i]=rev[i>>1]>>1|(i&1)*(len>>1);
	G[0][0]=5,G[1][0]=pw(5,TT-2);
	for (int T=0;T<2;T++)
	for (int i=1;i<len;i<<=1){
		G[T][i]=1;long long now=pw(G[T][0],(TT-1>>1)/i);
		for (int j=1;j<i;j++) G[T][i|j]=G[T][i|j-1]*now%TT;
	}
	for (int i=1,x,y;i<n;i++)
	x=read(),y=read(),add_e(x,y),add_e(y,x);
	DFS(1,-1);
	for (int i=1;i<=n;i++) a[i]=(a[i]+TT)*fct[i]%TT;
	for (int i=0;i<n;i++) b[i]=ifct[i];
	rsv(a);NTT(a,0),NTT(b,0);
	for (int i=0;i<len;i++) c[i]=a[i]*b[i]%TT;
	NTT(c,1);rsv(c);
	for (int i=1;i<=n;i++) printf("%lld\n",c[i]*ifct[i]%TT);
	return 0;
}