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<iostream>
#include<algorithm>
#include<vector>
#define pb push_back
using namespace std;
const int N=2e5+10,mod=924844033;
int n,siz[N],num[N],inv[N],jc[N],ans;
int r[N<<2],l,tt,A[N<<2],B[N<<2],w[N<<2];
vector<int> S[N];
int ksm(int x,int k)
{
	int s=1;for(;k;k>>=1,x=1ll*x*x%mod)
				if(k&1) s=1ll*s*x%mod;return s;
}
int C(int n,int k) {return 1ll*jc[n]*inv[k]%mod*inv[n-k]%mod;}
void NTT(int *P,int op)
{
	for(int i=1;i<l;i++) if(r[i]>i) swap(P[i],P[r[i]]);
	for(int i=1;i<l;i<<=1)
	{
		int W=ksm(5,(mod-1)/(i<<1));
		if(op<0) W=ksm(W,mod-2);w[0]=1;
		for(int j=1;j<i;j++) w[j]=1ll*w[j-1]*W%mod;
		for(int j=0,p=i<<1;j<l;j+=p)
			for(int k=0;k<i;k++)
			{
				int X=P[j+k],Y=1ll*P[j+k+i]*w[k]%mod;
				P[j+k]=(X+Y)%mod;P[j+k+i]=(mod+(X-Y)%mod)%mod;
			}
	}
}
void dfs1(int x,int fr)
{
	siz[x]=1;
	for(auto R:S[x]) if(R!=fr) dfs1(R,x),siz[x]+=siz[R];
}
void dfs2(int x,int fr)
{
	siz[x]+=siz[fr];
	for(auto R:S[x]) num[siz[R]]++;
	for(auto R:S[x]) if(R!=fr) siz[x]-=siz[R],dfs2(R,x),siz[x]+=siz[R];
	siz[x]-=siz[fr];
}
int main()
{
	cin>>n;
	for(int i=1,x,y;i<n;i++)
		cin>>x>>y,S[x].pb(y),S[y].pb(x);
	dfs1(1,0);dfs2(1,0);
	jc[0]=inv[0]=1;
	for(int i=1;i<=n;i++) jc[i]=1ll*jc[i-1]*i%mod;
	inv[n]=ksm(jc[n],mod-2);
	for(int i=n-1;i>=1;i--) inv[i]=1ll*inv[i+1]*(i+1)%mod;
	for(l=1;l<=n*2;l<<=1) tt++;tt--;
	for(int i=0;i<l;i++) r[i]=(r[i>>1]>>1)|((i&1)<<tt);
	for(int i=0;i<=n;i++) A[i]=1ll*jc[i]*num[i]%mod,B[i]=inv[i];
	reverse(B+0,B+n+1);NTT(A,1);NTT(B,1);
	for(int i=0;i<l;i++) A[i]=1ll*A[i]*B[i]%mod;NTT(A,-1);
	for(int i=0,inv=ksm(l,mod-2);i<l;i++) A[i]=1ll*A[i]*inv%mod;
	for(int i=1;i<=n;i++) printf("%lld\n",(mod+1ll*n*C(n,i)%mod-1ll*inv[i]*A[n+i]%mod)%mod);
}