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<iostream>
#include<cstring>
#include<algorithm>

#define N 600050
#define LL long long
const int mod=924844033,G1=5,G2=(3ll*mod+1)/G1;
using namespace std;
int n;
int he[N],ne[N<<2],to[N<<2],tot;
inline void addedge(int x,int y){
	to[++tot]=y;ne[tot]=he[x];he[x]=tot;
}
inline int qpow(int d,int z){
	int ret=1;
	for(;z;z>>=1,d=1ll*d*d%mod)
		if(z&1)ret=1ll*ret*d%mod;
	return ret;
}
int inc[N],inv[N];
inline int C(int n,int m){
	return 1ll*inc[n]*inv[m]*inv[n-m]%mod;
}
inline void init(int n){
	inc[0]=inv[0]=1;
	for(int i=1;i<=n;++i)inc[i]=1ll*inc[i-1]*i%mod;
	inv[n]=qpow(inc[n],mod-2);
	for(int i=n-1;i;--i)inv[i]=1ll*inv[i+1]*(i+1)%mod;
}
int sz[N],a[N],b[N];
inline void dfs(int g,int fa){
	sz[g]=1;
	for(int i=he[g];i;i=ne[i])
		if(to[i]!=fa)dfs(to[i],g),sz[g]+=sz[to[i]];
	if(fa)(a[sz[g]]+=inc[sz[g]])%=mod,(a[n-sz[g]]+=inc[n-sz[g]])%=mod;
}
inline void ntt(int *a,int n,int tag){
	static int r[N],cnt;cnt=-2;
	for(int i=n;i;i>>=1,++cnt);
	for(int i=0;i<n;++i){
		r[i]=(r[i>>1]>>1)|((i&1)<<cnt);
		if(r[i]>i)swap(a[i],a[r[i]]);
	}
	for(int i=1;i<n;i<<=1){
		int u=qpow(tag==1?G1:G2,mod/i/2),w=1;
		for(int j=0,I=i<<1;j<n;j+=I,w=1){
			for(int k=0;k<i;++k,w=1ll*w*u%mod){
				int t=1ll*w*a[j+k+i]%mod;
				a[j+k+i]=(a[j+k]+mod-t)%mod;
				a[j+k]=(a[j+k]+t)%mod;
			}
		}
	}
	if(tag==1)return;
	const LL iv=qpow(n,mod-2);
	for(int i=0;i<n;++i)a[i]=iv*a[i]%mod;
}

int main(){
	cin>>n;init(n+n);
	for(int i=1,x,y;i<n;++i)
		scanf("%d%d",&x,&y),addedge(x,y),addedge(y,x);
	dfs(1,0);
	for(int i=0;i<=n;++i)b[n-i]=inv[i];
	int len=1;while(len<=n)len<<=1;len<<=1;
	ntt(a,len,1);ntt(b,len,1);
	for(int i=0;i<len;++i)a[i]=1ll*a[i]*b[i]%mod;
	ntt(a,len,-1);
	for(int i=1;i<=n;++i)a[i]=a[i+n];
	for(int i=1;i<=n;++i){
		a[i]=1ll*inv[i]*(1ll*inc[n]*inv[n-i]%mod*n%mod+mod-a[i])%mod;
		printf("%d\n",(a[i])%mod);
	}
	return 0;
}