Problem: 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

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define fo(i,a,b) for(i=a;i<=b;i++)

using namespace std;

const int maxn=2e5+5,G=5,md=924844033;

int f[maxn*4],F[maxn],b[maxn*4],t[maxn*4],size[maxn];
int w[maxn*4],_w[maxn*4],cc[maxn],inv[maxn];
int fi[maxn],ne[maxn*2],dui[maxn*2],qc[maxn];
int i,j,k,l,m,n,now,x,y,N,ans;

void add(int x,int y){
	if (fi[x]==0) fi[x]=++now; else ne[qc[x]]=++now;
	dui[now]=y; qc[x]=now;
}
int quickmi(int x,int y){
	int t1=1;
	while (y){
		if ((y & 1)==1) t1=1ll*t1*x%md;
		x=1ll*x*x%md;
		y=y/2;
	}
	return t1;
}
void dfs(int x,int y){
	size[x]=1;
	for(int i=fi[x];i;i=ne[i]) if (dui[i]!=y){
		dfs(dui[i],x);
		F[size[dui[i]]]++; F[n-size[dui[i]]]++;
		size[x]=size[x]+size[dui[i]];
	}
}
void Ge_N(){
	N=1;
	while (N<=n) N=N*2;
	N=N*2;
}
void Ge_W(){
	w[0]=w[N]=1;
	int t=quickmi(G,(md-1)/N);
	fo(i,1,N-1) w[i]=1ll*w[i-1]*t%md;
	fo(i,0,N) _w[i]=w[N-i];
}
int reversal(int x){
	int t=0;
	for(int i=1;i<N;i<<=1,x>>=1) t=(t<<1)+(1&x);
	return t;
}
void DFT(int *x,int *w){
	int i,j,k,m,tmp;
	fo(i,0,N-1) t[i]=x[i];
	fo(i,0,N-1) x[reversal(i)]=t[i];
	for(i=2,m=1;i<=N;i<<=1,m<<=1){
		for(j=0;j<N;j+=i){
			fo(k,0,m-1){
				tmp=1ll*x[j+k+m]*w[N/i*k]%md;
				x[j+k+m]=(x[j+k]-tmp+md)%md;
				x[j+k]=(x[j+k]+tmp)%md;
			}
		}
	}
}
int C(int y,int x){
	return 1ll*inv[x]*inv[y-x]%md*cc[y]%md;
}
int main(){
//	freopen("t4.in","r",stdin);
	scanf("%d",&n);
	fo(i,1,n-1){
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}
	cc[0]=1; inv[0]=1;
	fo(i,1,n){
		cc[i]=1ll*cc[i-1]*i%md;
		inv[i]=quickmi(cc[i],md-2);
	}
	dfs(1,0);
	l=1; b[0]=1;
	fo(i,1,n){
		l=1ll*l*i%md;
		f[n-i]=1ll*F[i]*l%md;
		b[i]=quickmi(l,md-2);
	}
	Ge_N();
	Ge_W();
	DFT(f,w);
	DFT(b,w);
	fo(i,0,N) f[i]=1ll*f[i]*b[i]%md;
	DFT(f,_w);
	l=quickmi(N,md-2);
	fo(i,0,n) f[i]=1ll*f[i]*l%md;
	l=1;
	fo(i,1,n){
		l=1ll*l*i%md;
		ans=(1ll*C(n,i)*n-1ll*f[n-i]*inv[i]%md+md)%md;
		printf("%d\n",ans);
	}
	return 0;
}