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;
#define ref(i,x,y)for(int i=x;i<=y;++i)
#define SZ(x) ((int)x.size())
typedef vector<int> vec;typedef long long LL;
const int N=1<<19,M=924844033,G=5;
int n,s[N],f[N],F[N],r[N],t[N];vec A,B,e[N];
void dfs(int f,int x){
	ref(i,0,SZ(e[x])-1){int y=e[x][i];if(y!=f)dfs(x,y),s[x]+=s[y];}
	if(x>1)A[s[x]]++,A[n-s[x]]++;
}
int mi(int a,int b,int s=1){
	for(;b;b>>=1,a=(LL)a*a%M)if(b&1)s=(LL)s*a%M;
	return s;
}
void esbs(int L){
	ref(i,1,L-1)r[i]=r[i>>1]>>1|((i&1)?(L>>1):0);
	for(int i=1;i<L;i<<=1)
		for(int j=0,W=1,w=mi(G,(M-1)/(i<<1));j<i;++j,W=(LL)W*w%M)t[i+j]=W;
}
void dft(vec&a){
	int L=SZ(a);
	for(int i=1;i<L;++i)if(r[i]<i)swap(a[r[i]],a[i]);
	for(int i=1;i<L;i<<=1)
		for(int j=0;j<L;j+=(i<<1))
			for(int k=0;k<i;++k){
				int w=(LL)a[i+j+k]*t[i+k]%M;
				a[i+j+k]=a[j+k]-w;if(a[i+j+k]<0)a[i+j+k]+=M;
				a[j+k]+=w;if(a[j+k]>=M)a[j+k]-=M;
			}
}
void idft(vec&a){
	dft(a);int L=SZ(a),w=mi(L,M-2);
	ref(i,1,(L>>1)-1)swap(a[i],a[L-i]);
	ref(i,0,L-1)a[i]=(LL)a[i]*w%M;
}
int main(){
	cin>>n;
	ref(i,2,n){int x,y;cin>>x>>y;e[x].push_back(y);e[y].push_back(x);}
	f[0]=1;ref(i,1,n)s[i]=1,f[i]=(LL)f[i-1]*i%M;
	A.resize(n);dfs(0,1);ref(i,1,n-1)A[i]=(LL)A[i]*f[i]%M;
	B.resize(n+1);ref(i,0,n)B[n-i]=F[i]=mi(f[i],M-2);
	int l=SZ(A)+SZ(B)-1,L=1;for(;L<l;L<<=1);esbs(L);
	A.resize(L);B.resize(L);dft(A);dft(B);
	ref(i,0,L-1)A[i]=(LL)A[i]*B[i]%M;idft(A);A.resize(n+n+1);
	ref(i,1,n)cout<<(M+(LL)n*f[n]%M*F[i]%M*F[n-i]%M-(LL)F[i]*A[n+i]%M)%M<<endl;
}