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<bits/stdc++.h>

using namespace std;

typedef long long LL;
#define N 600000

const LL mod=924844033;
LL n,x,y,fac[N],ifac[N],rev[N],a[N],b[N],head[N],tot,sz[N];
struct edge{LL v,nxt;}e[N];

void add(LL x,LL y){
	e[++tot].v=y; e[tot].nxt=head[x]; head[x]=tot;
}

void dfs(LL u,LL fa){
	sz[u]+=1;
	for (LL i=head[u],v;i;i=e[i].nxt)
		if ((v=e[i].v)!=fa){
			dfs(v,u);
			sz[u]+=sz[v];
		}
	if (fa){++a[sz[u]]; ++a[n-sz[u]];}
}

LL pw(LL x,LL k){
	if (k==0) return 1;
	LL tmp=pw(x*x%mod,k>>1);
	return k&1?tmp*x%mod:tmp;
}

LL inv(LL x){
	return x==1?1:(mod-mod/x)*inv(mod%x)%mod;
}

void ntt_init(LL &p){
	LL k=0; for (;1<<k<p;++k);
	p=1<<k; rev[0]=0;
	for (LL i=1;i<p;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(k-1));
}

void ntt(LL f[],LL n){
	for (LL i=0;i<n;++i) if (i<rev[i]) swap(f[i],f[rev[i]]);
	for (LL i=0;1<<i<n;++i){
		LL t=1<<i,wn=pw(5,(mod-1)/t/2);
		for (LL j=0;j<n;j+=t<<1){
			LL *f1=f+j,*f2=f+j+t,w=1;
			for (LL k=0;k<t;++k){
				LL tmp=f2[k]*w%mod;
				f2[k]=(f1[k]-tmp+mod)%mod;
				f1[k]=(f1[k]+tmp)%mod;
				w=w*wn%mod;
			}
		}
	}
}

int main(){
	scanf("%lld",&n);
	fac[0]=1; for (LL i=1;i<=n;++i) fac[i]=fac[i-1]*i%mod;
	ifac[n]=inv(fac[n]); for (LL i=n-1;~i;--i) ifac[i]=ifac[i+1]*(i+1)%mod;
	for (LL i=1;i<n;++i){
		scanf("%lld%lld",&x,&y);
		add(x,y); add(y,x);
	}
	dfs(1,0);
	for (LL i=0;i<=n;++i){a[i]=a[i]*fac[i]%mod; b[i]=ifac[i];}
	for (LL i=0;i<=n/2;++i) swap(a[i],a[n-i]);
	LL p=n*2+1; ntt_init(p); ntt(a,p); ntt(b,p);
	for (LL i=0;i<p;++i) a[i]=a[i]*b[i]%mod;
	ntt(a,p); reverse(a+1,a+p);
	for (LL i=1,ii=inv(p);i<=n;++i){
		LL t=a[n-i]*ii%mod*ifac[i]%mod;
		b[i]=(n*fac[n]%mod*ifac[i]%mod*ifac[n-i]%mod-t+mod)%mod;
	}
	for (LL i=1;i<=n;++i) printf("%lld\n",b[i]);
	
	return 0;
}
			