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 <bits/stdc++.h>
using namespace std;
const int N=1<<19,mod=924844033;
int read(){
	int x=0;
	char ch=getchar();
	while (!isdigit(ch))
		ch=getchar();
	while (isdigit(ch))
		x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return x;
}
int Pow(int x,int y){
	int ans=1;
	for (;y;y>>=1,x=1LL*x*x%mod)
		if (y&1)
			ans=1LL*ans*x%mod;
	return ans;
}
int n;
vector <int> e[N];
int size[N],tax[N];
void dfs(int x,int pre){
	size[x]=1;
	for (auto y : e[x])
		if (y!=pre){
			dfs(y,x);
			size[x]+=size[y];
		}
	tax[size[x]]++;
	tax[n-size[x]]++;
}
int w[N],R[N],A[N],B[N];
void FFT(int a[N],int n){
	for (int i=0;i<n;i++)
		if (R[i]<i)
			swap(a[i],a[R[i]]);
	for (int t=n>>1,d=1;d<n;d<<=1,t>>=1)
		for (int i=0;i<n;i+=(d<<1))
			for (int j=0;j<d;j++){
				int tmp=1LL*w[t*j]*a[i+j+d]%mod;
				a[i+j+d]=(a[i+j]-tmp+mod)%mod;
				a[i+j]=(a[i+j]+tmp)%mod;
			}
}
void Mul(){
	int m,d;
	for (d=0,m=1;m<n*2;m<<=1,d++);
	w[0]=1,w[1]=Pow(5,(mod-1)/m);
	for (int i=2;i<m;i++)
		w[i]=1LL*w[i-1]*w[1]%mod;
	for (int i=0;i<m;i++)
		R[i]=(R[i>>1]>>1)|((i&1)<<(d-1));
	FFT(A,m);
	FFT(B,m);
	for (int i=0;i<m;i++)
		A[i]=1LL*A[i]*B[i]%mod;
	w[1]=Pow(w[1],mod-2);
	for (int i=2;i<m;i++)
		w[i]=1LL*w[i-1]*w[1]%mod;
	FFT(A,m);
	int inv=Pow(m,mod-2);
	for (int i=0;i<m;i++)
		A[i]=1LL*A[i]*inv%mod;
}
int Fac[N],Inv[N];
int C(int n,int m){
	if (m<0||m>n)
		return 0;
	return 1LL*Fac[n]*Inv[m]%mod*Inv[n-m]%mod;
}
int main(){
	n=read();
	for (int i=1;i<=n;i++)
		e[i].clear();
	for (int i=1;i<n;i++){
		int a=read(),b=read();
		e[a].push_back(b);
		e[b].push_back(a);
	}
	memset(tax,0,sizeof tax);
	dfs(1,0);
	Fac[0]=Inv[0]=1;
	for (int i=1;i<=n;i++){
		Fac[i]=1LL*Fac[i-1]*i%mod;
		Inv[i]=1LL*Inv[i-1]*Pow(i,mod-2)%mod;
	}
	for (int i=1;i<n;i++)
		A[i]=1LL*tax[i]*Fac[i]%mod;
	for (int i=0;i<=n;i++)
		B[i]=Inv[n-i];
	Mul();
	for (int i=1;i<=n;i++)
		A[i+n]=1LL*A[i+n]*Inv[i]%mod;
	for (int i=1;i<=n;i++)
		printf("%lld\n",(1LL*C(n,i)*n-A[i+n]+mod)%mod);
	return 0;
}

/*

C(n,k) - ΣC(size[son],k)   -   C(n-size[x],k)

多项式大小 小于子树size

问题变成了
ans_i = Σ C(a[j], i) = a[j]! / i! / (a[j]-i)! 
	  a[j] + (a[j]-i) = i

令 A[i] =  i! 
令 B[i] = 1/C[i-n] 
令 C[i] =  (-i)!  (i<=0)  
	    =  0	  (i>0)   

ans[i] = Σ A[j] * B[i-j]  

*/