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;
const int N=1e6+7,p=924844033; bool fl[N];
int n,k,lim,r[N],len[N],siz[N],ans[N],sum[N]; vector<int>v[N]; long long fac[N],ifac[N];
inline int read(){
	int num=0; char g=getchar(); while(g<48||57<g) g=getchar();
	while(47<g&&g<58) num=(num<<1)+(num<<3)+g-48,g=getchar(); return num;
}
inline long long pows(long long u,int v){
	long long ans=1; while(v>0) {if(v&1) ans=ans*u%p; u=u*u%p,v=v>>1;} return ans;
}
inline void find(int u){
	fl[u]=1,siz[u]=1;
	for(int i=0;i<len[u];i++){
		int x=v[u][i]; if(!fl[x]) find(x),ans[siz[x]]--,siz[u]+=siz[x];
	}
	ans[n-siz[u]]--,ans[n]++;
}
inline void NTT(int *s,int x){
    for(int i=0;i<lim;i++) if(r[i]<i) swap(s[i],s[r[i]]);
    for(int d=1;d<lim;d=d<<1){
        int w=pows(x,(p-1)/(d<<1));
        for(int i=0,r=d<<1;i<lim;i+=r){
            long long wn=1;
            for(int j=0;j<d;j++,wn=(wn*w)%p){
                int a=s[i+j],b=(s[i+j+d]*wn)%p;
                s[i+j]=(a+b)%p,s[i+j+d]=(a-b+p)%p;
            }
        }
    }
}
int main(){
	n=read(); int u,t; fac[0]=ifac[0]=1;
	for(int i=1;i<n;i++) u=read(),t=read(),v[u].push_back(t),v[t].push_back(u),len[u]++,len[t]++; find(1);
	for(int i=1;i<=n;i++) fac[i]=fac[i-1]*i%p,ifac[i]=pows(fac[i],p-2);
	for(int i=0;i<=n;i++) ans[i]=(ans[i]+p)*fac[i]%p;
	for(int i=0;i<=n;i++) sum[n-i]=ifac[i]; k=0,lim=1;
	while(lim<=(n<<1)) lim=lim<<1,k++;
	for(int i=0;i<lim;i++) r[i]=(r[i>>1]>>1)|((i&1)<<(k-1));
	NTT(ans,5),NTT(sum,5);
	for(int i=0;i<lim;i++) ans[i]=1ll*ans[i]*sum[i]%p; NTT(ans,pows(5,p-2));
	for(int i=1;i<=n;i++) printf("%d\n",ans[n+i]*ifac[i]%p*pows(lim,p-2)%p);
	return 0;
}