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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

const int N=1000005;
const int mod=924844033;

struct E{
	int to,next;
}mem[N];
int n,num,x,y,l,il;
int head[N],b[N],a[N],f[N],sz[N],g[N],fac[N],inv[N];

void add(int x,int y){
	num++;
	mem[num].to=y; mem[num].next=head[x];
	head[x]=num;
}

int quickpow(int x,int y){
	int s=1;
	for (;y;y>>=1,x=1ll*x*x%mod)
		if (y&1) s=1ll*s*x%mod;
	return s;
}

void init(){
	int i;
	fac[0]=inv[0]=1;
	for (i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mod;
	inv[n]=quickpow(fac[n],mod-2);
	for (i=n-1;i;i--) inv[i]=1ll*inv[i+1]*(i+1)%mod;
}

void dfs(int k,int pre){
	int j,u;
	sz[k]=1;
	for (j=head[k];j;j=mem[j].next){
		u=mem[j].to;
		if (u==pre) continue;
		dfs(u,k);
		sz[k]+=sz[u]; g[sz[u]]--;
	}
	if (k!=1) g[n-sz[k]]--;
}

void ntt(int z[],int b){
	int s,x,y,w,wd,i,j,k;
	for (i=0;i<l;i++){
		s=0;
		for (j=1,k=l>>1;j<l;j<<=1,k>>=1) if (i&j) s+=k;
		if (i<s) swap(z[i],z[s]);
	}
	for (i=1;i<l;i<<=1){
		wd=quickpow(5,(mod-1)/(i<<1))%mod; 
		if (b==-1) wd=quickpow(wd,mod-2);
		for (int j=0;j<l;j+=(i<<1)){
			w=1;
			for (int k=0;k<i;k++){
				x=z[j+k]; y=1ll*z[j+k+i]*w%mod;
				z[j+k]=(x+y)%mod;
				z[j+k+i]=(x-y+mod)%mod;
				w=1ll*w*wd%mod;
			}
		}
	} 
}

int main(){
	int i;
	scanf("%d",&n);
	for (i=1;i<n;i++){
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}
	init();
	dfs(1,0);
	for (l=1;l<n+n+2;l<<=1);
	g[n]=n;
	for (i=0;i<=n;i++) a[i]=1ll*(g[n-i]+mod)*fac[n-i]%mod;
	for (i=0;i<=n;i++) b[i]=inv[i];
	ntt(a,1);
	ntt(b,1);
	for (i=0;i<l;i++) f[i]=1ll*a[i]*b[i]%mod;
	ntt(f,-1);
	il=quickpow(l,mod-2);
	for (i=0;i<l;i++) f[i]=(1ll*f[i]*il%mod+mod)%mod;
	for (i=1;i<=n;i++) printf("%d\n",1ll*inv[i]*f[n-i]%mod);
	return 0;
}