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;
#define N 1000000
#define LL long long
#define mod 924844033
#define rep(i,j,k) for(i=j;i<=k;++i)
#define down(i,j,k) for(i=j;i>=k;--i)
const int g=5;
struct E{
	int to,nxt;
}ed[N];
int n,i,j,k,top,u,v,len,bit,invg,invn,wn,w;
int head[N],sz[N],cnt[N],fac[N],inv[N],a[N],b[N],rev[N],omg[N][2];
void read(int &p){
	p=0; char x=getchar();
	while(x<'0' || x>'9') x=getchar();
	while(x>='0'&&x<='9'){p=p*10+x-'0'; x=getchar();}
}
int quick_mi(int a,int b){
	int sum=1;
	for(;b;b>>=1){
		if(b&1) sum=(sum*1LL*a)%mod;
		a=(a*1LL*a)%mod;
	}
	return sum;
}
void add(int u,int v){
	top++; ed[top]=(E){v,head[u]}; head[u]=top;
}
void dfs(int x,int fa){
	int i;
	sz[x]=1;
	for(i=head[x];i;i=ed[i].nxt)
	if(ed[i].to!=fa){
		dfs(ed[i].to,x);
		sz[x]+=sz[ed[i].to];
	}
	if(x!=1){
		cnt[sz[x]]++;
		cnt[n-sz[x]]++;
	}
}
void init(int n){
	len=1; bit=0;
	while(len<n) len<<=1,bit++;
	invn=quick_mi(len,mod-2); 
	invg=quick_mi(g,mod-2);
	rep(i,0,bit){
		omg[1<<i][0]=quick_mi(g,(mod-1)/(1<<i));
		omg[1<<i][1]=quick_mi(invg,(mod-1)/(1<<i));
	}
	rep(i,0,len-1) rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1));
}
void ntt(int *a,int flag){
	rep(i,0,len-1)
	if(i<rev[i]) swap(a[i],a[rev[i]]);
	for(i=2;i<=len;i<<=1){
		wn=omg[i][flag];
		for(j=0;j<len;j+=i){
			w=1;
			rep(k,0,(i>>1)-1){
				int t=(w*1LL*a[j+k+(i>>1)])%mod;
				a[j+k+(i>>1)]=(((a[j+k]-t)%mod)+mod)%mod;
				a[j+k]=(a[j+k]+t)%mod;
				w=(w*1LL*wn)%mod;
			}
		}
	}
	if(flag)
	rep(i,0,len-1) a[i]=(a[i]*1LL*invn)%mod;
}
LL C(int n,int m){
	return (((fac[n]*1LL*inv[n-m])%mod)*1LL*inv[m])%mod;
}
int main(){
	read(n);
	rep(i,1,n-1){
		read(u); read(v);
		add(u,v); add(v,u);
	}
	fac[0]=1;
	rep(i,1,n) fac[i]=(fac[i-1]*1LL*i)%mod;
	inv[n]=quick_mi(fac[n],mod-2);
	down(i,n-1,0) b[n-i]=inv[i]=(inv[i+1]*1LL*(i+1LL))%mod;
	dfs(1,0);
	rep(i,1,n) a[i]=(cnt[i]*1LL*fac[i])%mod;
	init(n+n+2);
	ntt(a,0);
	ntt(b,0);
	rep(i,0,len-1) a[i]=(a[i]*1LL*b[i])%mod;
	ntt(a,1);
	rep(i,1,n){
		LL t=(((n*1LL*C(n,i))%mod)-((inv[i]*1LL*a[n+i])%mod))%mod;
		if(t<0) t+=mod;
		printf("%lld\n",t);
	}
	return 0;
}