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>
#define mod 924844033
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
int pow(int x,int y){
	int ret=1;
	while(y){
		if(y&1)ret=1ll*ret*x%mod;
		x=1ll*x*x%mod;y>>=1;
	}
	return ret;
}
int fact[200010],ifact[200010];
int C(int n,int m){if(n<m)return 0;return 1ll*fact[n]*ifact[m]%mod*ifact[n-m]%mod;}
int fir[200010],dis[400010],nxt[400010],id;
void link(int a,int b){nxt[++id]=fir[a],fir[a]=id,dis[id]=b;}
int siz[200010];
void dfs(int x){
	siz[x]=1;
	for(int i=fir[x];i;i=nxt[i]){
		if(siz[dis[i]])continue;
		dfs(dis[i]);
		siz[x]+=siz[dis[i]];
	}
}
int A[524341],B[524341],rev[524341],N,lg;
const int G=19260817,iG=pow(G,mod-2);
void ntt(int*A,int N,int t){
	for(int i=0;i<N;++i)if(i<rev[i])std::swap(A[i],A[rev[i]]);
	for(int o=1;o<N;o<<=1){
		int W=pow(t?G:iG,(mod-1)/(o<<1));
		for(int*p=A;p!=A+N;p+=o<<1)
			for(int i=0,w=1;i<o;++i,w=1ll*w*W%mod){
				int t=1ll*p[i+o]*w%mod;
				p[i+o]=(p[i]-t+mod)%mod;
				p[i]=(p[i]+t)%mod;
			}
	}
	if(!t){
		int iN=pow(N,mod-2);
		for(int i=0;i<N;++i)A[i]=1ll*A[i]*iN%mod;
	}
}
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	int n=gi(),a,b;
	fact[0]=1;for(int i=1;i<=n;++i)fact[i]=1ll*fact[i-1]*i%mod;
	ifact[n]=pow(fact[n],mod-2);for(int i=n;i;--i)ifact[i-1]=1ll*ifact[i]*i%mod;
	for(int i=1;i<n;++i)a=gi(),b=gi(),link(a,b),link(b,a);
	dfs(1);
	for(int i=2;i<=n;++i)++A[n-siz[i]],++A[siz[i]];
	for(int i=1;i<=n;++i)A[i]=1ll*A[i]*fact[n-i]%mod;
	for(int i=0;i<=n;++i)B[i]=ifact[i];
	N=1;while(N<n*2+3)N<<=1,++lg;
	for(int i=0;i<N;++i)rev[i]=(rev[i>>1]>>1)|((i&1)<<lg-1);
	ntt(A,N,1),ntt(B,N,1);
	for(int i=0;i<N;++i)A[i]=1ll*A[i]*B[i]%mod;
	ntt(A,N,0);
	for(int i=1;i<=n;++i)printf("%d\n",(1ll*n*C(n,i)-1ll*ifact[i]*A[n-i]%mod+mod)%mod);
	return 0;
}
