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 MaxN=2e5+5;
const int MaxM=(1<<19)+5;
const int Mod=924844033;

int N,M,InvM;
int Size[MaxN],Cnt[MaxN];
int R[MaxM],A[MaxM],B[MaxM],C[MaxM];
int Fac[MaxN],Inv[MaxN],Ans[MaxN];

vector<int> To[MaxN];

void DFS(int u,int fa){
	Size[u]=1;
	for(int v:To[u])
		if(v^fa){
			DFS(v,u);
			Size[u]+=Size[v];
		}
	if(u>1)
		Cnt[Size[u]]++;
}

int Pow(int base,int n){
	int ans=1;
	if(n<0)
		n+=Mod-1;
	for(;n;n>>=1){
		if(n&1)
			ans=1ll*ans*base%Mod;
		base=1ll*base*base%Mod;
	}
	return ans;
}

void NTT(int *a,int f){
	int i,j,k,wn,w,x,y;
	for(i=0;i<M;i++)
		if(i<R[i])
			swap(a[i],a[R[i]]);
	for(i=1;i<M;i<<=1){
		wn=Pow(5,f*(Mod-1)/(i<<1));
		for(j=0;j<M;j+=i<<1)
			for(w=1,k=0;k<i;k++){
				x=a[j+k],y=1llu*w*a[j+k+i]%Mod;
				a[j+k]=(x+y)%Mod,a[j+k+i]=(x-y+Mod)%Mod;
				w=1llu*w*wn%Mod;
			}
	}
}

int Binom(int n,int m){
	return n<m?0:1llu*Fac[n]*Inv[m]%Mod*Inv[n-m]%Mod;
}

int main(){
	int i,j,u,v;
	scanf("%d",&N);
	Fac[0]=Fac[1]=Inv[0]=Inv[1]=1;
	for(i=2;i<=N;i++){
		Fac[i]=1llu*Fac[i-1]*i%Mod;
		Inv[i]=1llu*(Mod-Mod/i)*Inv[Mod%i]%Mod;
	}
	for(i=2;i<=N;i++)
		Inv[i]=1llu*Inv[i-1]*Inv[i]%Mod;
	for(i=1;i<N;i++){
		scanf("%d%d",&u,&v);
		To[u].push_back(v);
		To[v].push_back(u);
	}
	DFS(1,0);
	for(i=0;i<=N;i++){
		A[i]=1llu*Fac[i]*(Cnt[i]+Cnt[N-i])%Mod;
		B[N-i]=Inv[i];
	}
	for(j=0,M=1;M<=2*N;j++,M<<=1);
	for(i=0;i<M;i++)
		R[i]=(R[i>>1]>>1)+((i&1)<<j-1);
	InvM=Pow(M,Mod-2);
	NTT(A,1),NTT(B,1);
	for(i=0;i<M;i++)
		C[i]=1llu*A[i]*B[i]%Mod;
	NTT(C,-1);
	for(i=1;i<=N;i++)
		printf("%d\n",(1llu*N*Binom(N,i)%Mod-1llu*C[N+i]*InvM%Mod*Inv[i]%Mod+Mod)%Mod);
	return 0;
}