Problem: 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

Solution (in CPP):

Solution:
#include<set>    
#include<map>    
#include<stack>    
#include<ctime>    
#include<cstdio>    
#include<queue>    
#include<cmath>    
#include<vector>    
#include<cstring>   
#include<climits>    
#include<iostream>   
#include<algorithm>   
using namespace std;
#define LL long long
int read(){   
    int f=1,x=0;char c=getchar();   
    while(c<'0'||'9'<c){if(c=='-')f=-1;c=getchar();}   
    while('0'<=c&&c<='9'){x=x*10+c-'0';c=getchar();}   
    return f*x;
}
#define MAXN int(5e6)
#define INF 0x3f3f3f3f
const LL P=924844033,g=5,rg=554906420;
LL Lim=1,lg2,a[MAXN+5],b[MAXN+5],rev[MAXN+5];
LL Pow(LL x,int y){
	LL ret=1;
	while(y){
		if(y&1) ret=(ret*x)%P;
		x=(x*x)%P;
		y>>=1;
	}
	return ret;
}
#define limit Lim
#define G g
#define Gi rg
void NTT(LL *A, int type) {
    for(int i=0;i<limit;i++) 
        if(i<rev[i]) swap(A[i],A[rev[i]]);
    for(int mid=1;mid<limit;mid <<=1) {  
        LL Wn=Pow(type==1?G:Gi,(P-1)/(mid<<1));
        for(int j=0;j<limit;j+=(mid<<1)) {
            LL w=1;
            for(int k=0;k<mid;k++,w=(w*Wn)%P) {
				int x=A[j+k],y=w*A[j+k+mid]%P;
				A[j+k]=(x+y)%P,A[j+k+mid]=(x-y+P)%P;
            }
        }
    }
}
vector<int> Gra[MAXN+5];
int N,siz[MAXN+5],cnt[MAXN+5];
void DFS(int u,int fa){
	siz[u]=1;
	for(int i=0;i<(int)Gra[u].size();i++){
		int v=Gra[u][i];
		if(v==fa) continue;
		DFS(v,u),siz[u]+=siz[v];
	}
	if(fa!=0)
		cnt[N]++,cnt[siz[u]]--,cnt[N-siz[u]]--;
	return ;
}
LL fac[MAXN+5],inv[MAXN+5];
int main(){
	N=read();
	fac[0]=1;
	for(int i=1;i<=N;i++)
		fac[i]=fac[i-1]*i%P;
	inv[N]=Pow(fac[N],P-2);
	for(int i=N-1;i>=0;i--)
		inv[i]=inv[i+1]*(i+1)%P;
	for(int i=1;i<N;i++){
		int u=read(),v=read();
		Gra[u].push_back(v);
		Gra[v].push_back(u);
	}
	DFS(1,0),cnt[N]=N;
	//
	int n=N,m=N;
	for(int i=0;i<=n;i++)
		a[i]=cnt[i]*fac[i]%P;
	for(int i=0;i<=m;i++)
		b[m-i]=inv[i];
	while(Lim<=n+m) Lim<<=1,lg2++;
	for(int i=0;i<Lim;i++)
		rev[i]=(rev[i>>1]>>1)|((i&1)<<(lg2-1));
	NTT(a,1),NTT(b,1);
	for(int i=0;i<Lim;i++)
		a[i]=(a[i]*b[i])%P;
	NTT(a,-1);
	LL inv_Lim=Pow(Lim,P-2);
	for(int i=0;i<=n+m;i++)
		a[i]=a[i]*inv_Lim%P;
	for(int i=1;i<=n;i++)
		printf("%lld\n",a[n+i]*inv[i]%P);
	return 0;
}//