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 pb push_back
using namespace std;
const int N=1000100;
const int mod=924844033;
const int g[2]={5,554906420};
int n,lim=1,pos[N],cnt[N],siz[N],f[N],r[N],T[N];
vector<int>V[N];
int qpw(int x,int n){
	int ret=1; for(;n;n>>=1,x=1ll*x*x%mod)
	if(n&1) ret=1ll*ret*x%mod; return ret;
}
void ini(int x){
	int l=0; while(lim<x) lim<<=1,++l;
	for(int i=0;i<lim;i++) pos[i]=pos[i>>1]>>1|(i&1)<<l-1;
}
void ntt(int A[],int op){
	for(int i=0;i<lim;i++) if(i<pos[i]) swap(A[i],A[pos[i]]);
	for(int d=1;d<lim;d<<=1) for(int l=d<<1,wn=qpw(g[op],(mod-1)/l),i=0;i<lim;i+=l)
		for(int j=0,w=1;j<d;j++,w=1ll*w*wn%mod){
			int a=A[i+j],b=1ll*w*A[i+j+d]%mod;
			A[i+j]=(a+b)%mod,A[i+j+d]=(a-b)%mod;
		}
	if(op) for(int i=0,k=qpw(lim,mod-2);i<lim;i++) A[i]=1ll*A[i]*k%mod;
}
void dfs(int x,int f){
	siz[x]=1;
	for(int i=0;i<V[x].size();i++){
		int t=V[x][i]; if(t==f) continue; dfs(t,x);
		++cnt[siz[t]], ++cnt[n-siz[t]], siz[x]+=siz[t];
	}
}
int C(int n,int m){return 1ll*f[n]*r[n-m]%mod*r[m]%mod;}
int main(){
	scanf("%d",&n),f[0]=1;
	for(int i=1;i<=n;i++) f[i]=1ll*f[i-1]*i%mod;
	r[n]=qpw(f[n],mod-2); for(int i=n;i>=1;i--) r[i-1]=1ll*r[i]*i%mod;
	for(int i=1,x,y;i<n;i++)scanf("%d%d",&x,&y),V[x].pb(y),V[y].pb(x);
	dfs(1,0); for(int i=0;i<=n;i++) cnt[i]=1ll*cnt[i]*f[i]%mod, T[i]=r[i]; reverse(cnt,cnt+n+1);
	ini(n<<1), ntt(cnt,0), ntt(T,0); for(int i=0;i<lim;i++) cnt[i]=1ll*cnt[i]*T[i]%mod; ntt(cnt,1);
	for(int i=1;i<=n;i++) printf("%lld\n",(mod+(1ll*n*C(n,i)-1ll*r[i]*cnt[n-i])%mod)%mod);
}
