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>
#define maxn 530005
#define rep(i,j,k) for(int i=(j),lim=(k);i<=lim;i++)
#define Clear(a,l,r) memset(a+(l),0,((r)-(l)+1)*(sizeof a[0]))
using namespace std;
const int mod = 924844033, inv2 = (mod+1)/2;
int w[maxn],WL,r[maxn],lg[maxn],inv[maxn],fac[maxn],invf[maxn];
int Pow(int a,int b){int s=1;for(;b;b>>=1,a=1ll*a*a%mod) b&1&&(s=1ll*s*a%mod); return s;}
void init(int n){
	w[0]=WL=inv[0]=inv[1]=fac[0]=fac[1]=invf[0]=invf[1]=1;
	while(WL<=n) WL<<=1; w[1]=Pow(5,(mod-1)/WL);
	for(int i=2;i<=WL;i++) w[i]=1ll*w[i-1]*w[1]%mod,lg[i]=lg[i>>1]+1;
	for(int i=2;i<=WL;i++) fac[i]=1ll*fac[i-1]*i%mod,inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod,invf[i]=1ll*invf[i-1]*inv[i]%mod;
}
int upd(int x){return x+(x>>31&mod);}
void NTT(int *a,int len,int flg){
	for(int i=0;i<len;i++) if(i<(r[i]=r[i>>1]>>1|(i&1?len>>1:0))) swap(a[i],a[r[i]]);
	for(int i=2,l=1,v;i<=len;l=i,i<<=1) for(int j=0,t=WL/i;j<len;j+=i) for(int k=j,o=0;k<j+l;k++,o+=t)
		a[k+l]=upd(a[k]-(v=1ll*w[flg^1?WL-o:o]*a[k+l]%mod)),a[k]=upd(a[k]+v-mod);
	if(flg^1) for(int i=0,Inv=mod-(mod-1)/len;i<len;i++) a[i]=1ll*a[i]*Inv%mod;
}
int n,siz[maxn],A[maxn],B[maxn];
vector<int>G[maxn];
void dfs(int u,int ff){siz[u]=1; for(int v:G[u]) if(v!=ff) dfs(v,u),siz[u]+=siz[v];}
int main()
{
	scanf("%d",&n),init(2*n);
	for(int i=1,x,y;i<n;i++) scanf("%d%d",&x,&y),G[x].push_back(y),G[y].push_back(x);
	dfs(1,0);
	for(int i=2;i<=n;i++) A[siz[i]]++,A[n-siz[i]]++;
	for(int i=0;i<n;i++) A[i]=1ll*A[i]*fac[i]%mod,B[i]=invf[n-1-i];
	NTT(A,WL,1),NTT(B,WL,1);
	for(int i=0;i<WL;i++) A[i]=1ll*A[i]*B[i]%mod;
	NTT(A,WL,-1);
	for(int i=1;i<=n;i++)
		printf("%d\n",upd(1ll*n*fac[n]%mod*invf[i]%mod*invf[n-i]%mod-1ll*A[n-1+i]*invf[i]%mod));
}